### Chapter 1: Mathematical Logic. State which of the following sentences is a statement. Justify your answer if it is a statement. Write down its truth value.

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

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

## 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

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

## 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

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

## 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

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

## (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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## 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

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

## Can you speak in English?

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

EXERCISE 1.2 [PAGE 6]

### 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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

Exercise 1.2 | Q 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.

EXERCISE 1.3PAGE 7

### 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

Exercise 1.3 | Q 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

Exercise 1.3 | Q 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

Exercise 1.3 | Q 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

Exercise 1.3 | Q 1.4 | Page 7

Write the negation of the following statement.

2 + 3 ≠ 5

2 + 3 = 5

#### EXERCISE 1.3Q 2.1  PAGE 7

Exercise 1.3 | Q 2.1 | Page 7

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

$5$

## √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

Exercise 1.3 | Q 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

Exercise 1.3 | Q 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.

EXERCISE 1.4PAGES 10 - 11

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

EXERCISE 1.4Q 1.1    PAGE 10
Exercise 1.4 | Q 1.1 | Page 10

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.

EXERCISE 1.4Q 1.2   PAGE 10
Exercise 1.4 | Q 1.2 | Page 10

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.

EXERCISE 1.4Q 1.3   PAGE 10
Exercise 1.4 | Q 1.3 | Page 10

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.

EXERCISE 1.4Q 1.4   PAGE 10
Exercise 1.4 | Q 1.4 | Page 10

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.

EXERCISE 1.4Q 1.5   PAGE 10
Exercise 1.4 | Q 1.5 | Page 10

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.

EXERCISE 1.4Q 1.6   PAGE 10
Exercise 1.4 | Q 1.6 | Page 10

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.

EXERCISE 1.4Q 2.1    PAGE 11
Exercise 1.4 | Q 2.1 | Page 11

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.

EXERCISE 1.4Q 2.2    PAGE 11
Exercise 1.4 | Q 2.2 | Page 11

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.

EXERCISE 1.4Q 2.3   PAGE 11
Exercise 1.4 | Q 2.3 | Page 11

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.

EXERCISE 1.4Q 2.4    PAGE 11
Exercise 1.4 | Q 2.4 | Page 11

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.

EXERCISE 1.4Q 2.5    PAGE 11
Exercise 1.4 | Q 2.5 | Page 11

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.

EXERCISE 1.4Q 3.1    PAGE 11
Exercise 1.4 | Q 3.1 | Page 11

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.

EXERCISE 1.4Q 3.2    PAGE 11

Exercise 1.4 | Q 3.2 | Page 11

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.

EXERCISE 1.4Q 3.3    PAGE 11
Exercise 1.4 | Q 3.3 | Page 11

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.

EXERCISE 1.4Q 3.4    PAGE 11
Exercise 1.4 | Q 3.4 | Page 11

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.

EXERCISE 1.4Q 3.5    PAGE 11
Exercise 1.4 | Q 3.5 | Page 11

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.

EXERCISE 1.4Q 3.6   PAGE 11
Exercise 1.4 | Q 3.6 | Page 11

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.

EXERCISE 1.4Q 4.1    PAGE 11
Exercise 1.4 | Q 4.1 | Page 11

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.

EXERCISE 1.4Q 4.2    PAGE 11
Exercise 1.4 | Q 4.2 | Page 11

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.

EXERCISE 1.4Q 4.3    PAGE 11
Exercise 1.4 | Q 4.3 | Page 11

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.

EXERCISE 1.4Q 5.1    PAGE 11
Exercise 1.4 | Q 5.1 | Page 11

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.

EXERCISE 1.4Q 5.2   PAGE 11
Exercise 1.4 | Q 5.2 | Page 11

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.

EXERCISE 1.4Q 5.3    PAGE 11
Exercise 1.4 | Q 5.3 | Page 11

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.

EXERCISE 1.4Q 5.4    PAGE 11
Exercise 1.4 | Q 5.4 | Page 11

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.

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

EXERCISE 1.5Q 1.1    PAGE 12

Exercise 1.5 | Q 1.1 | Page 12

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

x2 + 3x - 10 = 0

#### SOLUTION

∃ x ∈ N, such that x2 + 3x – 10 = 0

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

EXERCISE 1.5Q 1.2   PAGE 12
Exercise 1.5 | Q 1.2 | Page 12

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.

EXERCISE 1.5Q 1.3    PAGE 12
Exercise 1.5 | Q 1.3 | Page 12

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

n2 ≥ 1

#### SOLUTION

∀ n ∈ N, n2 ≥ 1

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

EXERCISE 1.5Q 1.4    PAGE 12
Exercise 1.5 | Q 1.4 | Page 12

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.

EXERCISE 1.5Q 1.5    PAGE 12
Exercise 1.5 | Q 1.5 | Page 12

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.

EXERCISE 1.5Q 1.6    PAGE 12
Exercise 1.5 | Q 1.6 | Page 12

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.

EXERCISE 1.5Q 2.1    PAGE 12
Exercise 1.5 | Q 2.1 | Page 12

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.

EXERCISE 1.5Q 2.2    PAGE 12
Exercise 1.5 | Q 2.2 | Page 12

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.

EXERCISE 1.5Q 2.3    PAGE 12
Exercise 1.5 | Q 2.3 | Page 12

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.

EXERCISE 1.5Q 2.4    PAGE 12
Exercise 1.5 | Q 2.4 | Page 12

If B = {2, 3, 5, 6, 7} determine the truth value of
∀ y ∈ B, such that y2 is negative.

#### SOLUTION

There is no y in B which satisfies y2 < 0.

∴ The given statement is false.

∴ It’s truth value is F.

EXERCISE 1.5Q 2.5    PAGE 12
Exercise 1.5 | Q 2.5 | Page 12

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 1.6 [PAGE 16]

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

EXERCISE 1.6Q 1.1   PAGE 16
Exercise 1.6 | Q 1.1 | 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
EXERCISE 1.6Q 1.2   PAGE 16
Exercise 1.6 | Q 1.2 | Page 16

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
EXERCISE 1.6Q 1.3   PAGE 16
Exercise 1.6 | Q 1.3 | Page 16

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
EXERCISE 1.6Q 1.4   PAGE 16
Exercise 1.6 | Q 1.4 | Page 16

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
EXERCISE 1.6Q 2.1   PAGE 16
Exercise 1.6 | Q 2.1 | Page 16

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.

EXERCISE 1.6Q 2.2   PAGE 16
Exercise 1.6 | Q 2.2 | Page 16

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.

EXERCISE 1.6Q 2.3   PAGE 16
Exercise 1.6 | Q 2.3 | Page 16

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.

EXERCISE 1.6Q 2.4   PAGE 16
Exercise 1.6 | Q 2.4 | Page 16

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.

EXERCISE 1.6Q 3.1   PAGE 16
Exercise 1.6 | Q 3.1 | Page 16

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.

EXERCISE 1.6Q 3.2   PAGE 16
Exercise 1.6 | Q 3.2 | Page 16

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.

EXERCISE 1.6Q 3.3   PAGE 16
Exercise 1.6 | Q 3.3 | Page 16

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.

EXERCISE 1.6Q 3.4   PAGE 16
Exercise 1.6 | Q 3.4 | Page 16

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.

EXERCISE 1.6Q 4.1   PAGE 16
Exercise 1.6 | Q 4.1 | Page 16

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.

EXERCISE 1.6Q 4.2   PAGE 16
Exercise 1.6 | Q 4.2 | Page 16

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.

EXERCISE 1.6Q 4.3   PAGE 16
Exercise 1.6 | Q 4.3 | Page 16

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.

EXERCISE 1.6Q 4.4   PAGE 16
Exercise 1.6 | Q 4.4 | Page 16

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.

EXERCISE 1.6Q 5.1   PAGE 16
Exercise 1.6 | Q 5.1 | Page 16

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.

EXERCISE 1.6Q 5.2   PAGE 16
Exercise 1.6 | Q 5.2 | Page 16

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.

EXERCISE 1.6Q 5.3   PAGE 16
Exercise 1.6 | Q 5.3 | Page 16

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.

EXERCISE 1.6Q 5.4   PAGE 16
Exercise 1.6 | Q 5.4 | Page 16

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 1.6Q 6.1   PAGE 16

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)

EXERCISE 1.6Q 6.2   PAGE 16
Exercise 1.6 | Q 6.2 | Page 16

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)

EXERCISE 1.6Q 6.3   PAGE 16
Exercise 1.6 | Q 6.3 | Page 16

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

EXERCISE 1.6Q 6.4   PAGE 16
Exercise 1.6 | Q 6.4 | Page 16

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 1.6Q 7.1   PAGE 16

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)

EXERCISE 1.6Q 7.2   PAGE 16
Exercise 1.6 | Q 7.2 | Page 16

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.

EXERCISE 1.6Q 7.3   PAGE 16
Exercise 1.6 | Q 7.3 | Page 16

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.

EXERCISE 1.6Q 7.4   PAGE 16
Exercise 1.6 | Q 7.4 | Page 16

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 1.7 [PAGE 17]

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

EXERCISE 1.7Q 1.1   PAGE 17
Exercise 1.7 | Q 1.1 | Page 17

Write the dual of the following:

(p ∨ q) ∨ r

#### SOLUTION

(p ∧ q) ∧ r

EXERCISE 1.7Q 1.2   PAGE 17
Exercise 1.7 | Q 1.2 | Page 17

Write the dual of the following:

~(p ∨ q) ∧ [p ∨ ~ (q ∧ ~ r)]

#### SOLUTION

~(p ∧ q) ∨ [p ∧ ~ (q ∨ ~ r)]

EXERCISE 1.7Q 1.3   PAGE 17
Exercise 1.7 | Q 1.3 | Page 17

Write the dual of the following:

p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r

#### SOLUTION

p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

EXERCISE 1.7Q 1.4   PAGE 17
Exercise 1.7 | Q 1.4 | Page 17

Write the dual of the following:

~(p ∧ q) ≡ ~ p ∨ ~ q

#### SOLUTION

~(p ∨ q) ≡ ~ p ∧ ~ q

EXERCISE 1.7Q 2.1   PAGE 17
Exercise 1.7 | Q 2.1 | Page 17

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.

EXERCISE 1.7Q 2.2   PAGE 17
Exercise 1.7 | Q 2.2 | Page 17

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.

EXERCISE 1.7Q 2.3   PAGE 17
Exercise 1.7 | Q 2.3 | Page 17

Write the dual statement of the following compound statement.

#### SOLUTION

EXERCISE 1.7Q 2.4   PAGE 17
Exercise 1.7 | Q 2.4 | Page 17

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.

EXERCISE 1.8 [PAGE 21]

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.8 [Page 21]

EXERCISE 1.8Q 1.1   PAGE 21
Exercise 1.8 | Q 1.1 | Page 21

Write the negation of the following statement.

All the stars are shining if it is night.

#### SOLUTION

Let q : All stars are shining.

p : It is night.

The given statement in symbolic form is p → q. It’s negation is ~ (p → q) ≡ p ∧ ~ q

∴ The negation of a given statement is ‘It is night and some stars are not shining’.

EXERCISE 1.8Q 1.2   PAGE 21
Exercise 1.8 | Q 1.2 | Page 21

Write the negation of the following statement.

∀ n ∈ N, n + 1 > 0

#### SOLUTION

∃ n ∈ N such that n + 1 ≤ 0.

EXERCISE 1.8Q 1.3   PAGE 21
Exercise 1.8 | Q 1.3 | Page 21

Write the negation of the following statement.

∃ n ∈ N, (n2 + 2) is odd number.

#### SOLUTION

∀ n ∈ N, (n2 + 2) is not odd number.

EXERCISE 1.8Q 1.4   PAGE 21
Exercise 1.8 | Q 1.4 | Page 21

Write the negation of the following statement.

Some continuous functions are differentiable.

#### SOLUTION

All continuous functions are not differentiable.

EXERCISE 1.8Q 2.1   PAGE 21
Exercise 1.8 | Q 2.1 | Page 21

Using the rules of negation, write the negation of the following:

(p → r) ∧ q

#### SOLUTION

~ [(p → r) ∧ q] ≡ ~(p → r) ∨ ~q  ....[Negation of conjunction]

≡ (p ∧ ~ r) ∨ ~q   .....[Negation of implication]

EXERCISE 1.8Q 2.2   PAGE 21
Exercise 1.8 | Q 2.2 | Page 21

Using the rules of negation, write the negation of the following:

~(p ∨ q) → r

#### SOLUTION

~[~(p ∨ q) → r] ≡ ~(p ∨ q) ∧ ~r  ....[Negation of implication]

≡ (~p ∧ ~q) ∧ ~r   .....[Negation of disjunction]

EXERCISE 1.8Q 2.3   PAGE 21
Exercise 1.8 | Q 2.3 | Page 21

Using the rules of negation, write the negation of the following:

(~p ∧ q) ∧ (~q ∨ ~r)

#### SOLUTION

~[(~p ∧ q) ∧ (~q ∨ ~r)]

≡ ~(~ p ∧ q) ∨ ~ (~ q ∨ ~r)    ...[Negation of conjunction]

≡ [~(~ p) ∨ ~ q] ∨ [~(~q) ∧ ~(~r)]  ...[Negation of conjunction and disjunction]

≡ (p ∨ ~q) ∨ (q ∨ r)      .....[Negation on negation]

EXERCISE 1.8Q 3.1   PAGE 21
Exercise 1.8 | Q 3.1 | Page 21

Write the converse, inverse, and contrapositive of the following statement.

If it snows, then they do not drive the car.

#### SOLUTION

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.

EXERCISE 1.8Q 3.2   PAGE 21
Exercise 1.8 | Q 3.2 | Page 21

Write the converse, inverse, and contrapositive of the following statement.

If he studies, then he will go to college.

#### SOLUTION

Let p : He studies.
q : He will go to college.

∴ The given statement is p → q.

Its converse is q → p.
If he will go to college then he studies.

Its inverse is ~p → ~q.
If he does not study then he will not go to college.

Its contrapositive is ~q → ~p.
If he will not go to college then he does not study.

EXERCISE 1.8Q 4.1   PAGE 21
Exercise 1.8 | Q 4.1 | Page 21

With proper justification, state the negation of the following.

(p → q) ∨ (p → r)

#### SOLUTION

~[(p → q) ∨ (p → r)]

≡ ~(p → q) ∧ ~(p → r)    ...[Negation of disjunction]

≡ (p ∧ ~ q) ∧ (p ∧ ~r)    ....[Negation of implication]

EXERCISE 1.8Q 4.2   PAGE 21
Exercise 1.8 | Q 4.2 | Page 21

With proper justification, state the negation of the following.

(p ↔ q) ∨ (~q → ~r)

#### SOLUTION

~[(p ↔ q) ∨ (~q → ~r)]

≡ ~(p ↔ q) ∧ (~q → ~r)        ....[Negation of disjunction]

≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ ~(~q → ~r)    ....[Negation of double implication]

≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ [~ q ∧ ~(~r)]    ....[Negation of implication]

≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ (~ q ∧ r)     ....[Negation of negation]

EXERCISE 1.8Q 4.3   PAGE 21
Exercise 1.8 | Q 4.3 | Page 21

With proper justification, state the negation of the following.

(p → q) ∧ r

#### SOLUTION

~[(p → q) ∧ r]

≡ ~ (p → q) ∨ ~ r    ....[Negation of conjunction]

≡ (p ∧ ~q) ∨ ~ r      ....[Negation of implication]

EXERCISE 1.9 [PAGE 22]

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.9 [Page 22]

EXERCISE 1.9Q 1.1   PAGE 22
Exercise 1.9 | Q 1.1 | Page 22

Without using truth table, show that

p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)

#### SOLUTION

L.H.S.

≡ p ↔ q

≡ (p → q) ∧ (q → p)

≡ (~p ∨ q) ∧ (~q ∨ p)

≡ [~ p ∧ (~ q ∨ p)] ∨ [q ∧ (~ q ∨ p)]     ....[Distributive law]

≡ [(~ p ∧ ~ q) ∨ (~ p ∧ p)] ∨ [(q ∧ ~ q) ∨ (q ∧ p)]     .....[Distributive Law]

≡ [(~ p ∧ ~ q) ∨ F] ∨ [F ∨ (q ∧ p)]  ....[Complement Law]

≡ (~ p ∧ ~ q) ∨ (q ∧ p)   ....[Identity Law]

≡ (p ∧ q) ∨ (~ p ∧ ~ q)    ....[Commutative Law]

≡ R.H.S.

EXERCISE 1.9Q 1.2   PAGE 22
Exercise 1.9 | Q 1.2 | Page 22

Without using truth table, show that

p ∧ [(~ p ∨ q) ∨ ~ q] ≡ p

#### SOLUTION

L.H.S.

≡ p ∧ [(~ p ∨ q) ∨ ~ q]

≡ p ∧ [(~ p ∨ (q ∨ ~ q)]     .....[Associative law]

≡ p ∧ (~ p ∨ T)       .....[Complement law]

≡ p ∧ T                 .....[Identity law]

≡ p                       .....[Identity law]

≡ R.H.S.

EXERCISE 1.9Q 1.3   PAGE 22
Exercise 1.9 | Q 1.3 | Page 22

Without using truth table, show that

~ [(p ∧ q) → ~ q] ≡ p ∧ q

#### SOLUTION

L.H.S.

≡ ~ [(p ∧ q) → ~ q]

≡ (p ∧ q) ∧ ~ (~ q)   ....[Negation of implication]

≡ (p ∧ q) ∧ q      .....[Negation of a negation]

≡ p ∧ (q ∧ q)     ....[Associative law]

≡ p ∧ q          .....[Identity law]

≡ R.H.S.

EXERCISE 1.9Q 1.4   PAGE 22
Exercise 1.9 | Q 1.4 | Page 22

Without using truth table, show that

~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p

#### SOLUTION

L.H.S.

≡ ~r → ~ (p ∧ q)

≡ ~(~ r) ∨ ~ (p ∧ q)       ....[p → q ≡ ~ p ∨ q]

≡ r ∨ ~(p ∧ q)             ....[Negation of negation]

≡ r ∨ (~p ∨ ~q)           ....[De Morgan’s law]

≡ ~p ∨ (~q ∨ r)         .....[Commutative and associative law]

≡ ~p ∨ (q → r)          ....[p → q ≡ ~ p ∨ q]

≡ (q → r) ∨ ~p            ......[Commutative law]

≡ ~[~ (q → r)] ∨ ~ p     ......[Negation of negation]

≡ [~ (q → r)] → ~ p        .....[p → q ≡ ~ p ∨ q]

= R.H.S.

EXERCISE 1.9Q 1.5   PAGE 22
Exercise 1.9 | Q 1.5 | Page 22

Without using truth table, show that

(p ∨ q) → r ≡ (p → r) ∧ (q → r)

#### SOLUTION

L.H.S.

≡ (p ∨ q) → r

≡ ~ (p ∨ q) ∨ r          ....[p → q → ~ p ∨ q]

≡ (~ p ∧ ~ q) ∨ r       ....[De Morgan’s law]

≡ (~ p ∨ r) ∧ (~ q ∨ r)       .....[Distributive law]

≡ (p → r) ∧ (q → r)             .....[p → q → ~ p ∨ q]

= R.H.S.

EXERCISE 1.9Q 2.1   PAGE 22
Exercise 1.9 | Q 2.1 | Page 22

Using the algebra of statement, prove that

[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p

#### SOLUTION

L.H.S.

= [p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p]

≡ [p ∧ (q ∨ r)] ∨ [(~ r ∧ ~ q)∧ p]     ...[Associative Law]

≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p]       ....[Commutative Law]

≡ [p ∧ (q ∨ r)] ∨ [~ (q ∨ r) ∧ p]         ....[De Morgan’s Law]

≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)]          .....[Commutative Law]

≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)]         ....[Distributive Law]

≡ p ∧ t          ......[Complement Law]

≡ p                .....[Identity Law]

= R.H.S.

EXERCISE 1.9Q 2.2   PAGE 22
Exercise 1.9 | Q 2.2 | Page 22

Using the algebra of statement, prove that

(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)

#### SOLUTION

L.H.S.

= (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q)

≡ (p ∧ q) ∨ [(p ∧ ~ q) ∨ (~ p ∧ ~ q)]    ....[Associative Law]

≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~ q ∧ ~ p)]    ....[Commutative Law]

≡ (p ∧ q) ∨ [~q ∧ (p ∨ ~ p)]      ....[Distributive Law]

≡ (p ∧ q) ∨ (~q ∧ t)      .....[Complement Law]

≡ (p ∧ q) ∨ (~q)            .....[Identity Law]

≡ (p ∨ ~ q) ∧ (q ∨ ~q)      .....[Distributive Law]

≡ (p ∨ ~ q) ∧ t              ....[Complement Law]

≡ p ∨ ~ q                   .....[Identity Law]

= R.H.S.

EXERCISE 1.9Q 2.3   PAGE 22
Exercise 1.9 | Q 2.3 | Page 22

Using the algebra of statement, prove that

#### SOLUTION

(p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q)

L.H.S.

= (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) ∨ (~ p ∧ q)      ....[Commutative law]

= R.H.S.

EXERCISE 1.10 [PAGES 22 - 27]

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.10 [Pages 22 - 27]

EXERCISE 1.10Q 1.1  PAGE 27
Exercise 1.10 | Q 1.1 | Page 27

Represent the truth of the following statement by the Venn diagram.

Some hardworking students are obedient.

#### SOLUTION

Let U : The set of all students.
H : The set of all hardworking students.
O : The set of all obedient students.

The above Venn diagram represents truth of the given statement, H ∩ O ≠ φ
EXERCISE 1.10Q 1.2  PAGE 27
Exercise 1.10 | Q 1.2 | Page 27

Represent the truth of the following statement by the Venn diagram.

No circles are polygons.

#### SOLUTION

Let U : The set of all closed geometrical figures in plane.
P : The set of all polygons
C : The set of all circles.

The above Venn diagram represents truth of the given statement, P ∩ C ≠ φ

EXERCISE 1.10Q 1.3  PAGE 27
Exercise 1.10 | Q 1.3 | Page 27

Represent the truth of the following statement by the Venn diagram.

All teachers are scholars and scholars are teachers.

#### SOLUTION

Let U : The set of all human beings.
T : The set of all teachers.
S : The set of all scholars

The above Venn diagram represents truth of the given statement, T = S

EXERCISE 1.10Q 1.4  PAGE 27
Exercise 1.10 | Q 1.4 | Page 22

Represent the truth of the following statement by the Venn diagram.

If a quadrilateral is a rhombus, then it is a parallelogram.

#### SOLUTION

Let U : The set of all quadrilaterals.
P : The set of all parallelograms.
R : The set of all rhombuses.

The above Venn diagram represents truth of the given statement, R ⊂ P.

EXERCISE 1.10Q 2.1  PAGE 27
Exercise 1.10 | Q 2.1 | Page 27

Draw a Venn diagram for the truth of the following statement.

Some share brokers are chartered accountants.

#### SOLUTION

Let U : The set of all human beings.
S : The set of all share brokers.
C : The set of all chartered accountants.

The above Venn diagram represents the truth of the given statement i.e., S ∩ C ≠ φ.

EXERCISE 1.10Q 2.2  PAGE 27
Exercise 1.10 | Q 2.2 | Page 27

Draw a Venn diagram for the truth of the following statement.

No wicket keeper is bowler, in a cricket team.

#### SOLUTION

Let U : The set of all human beings.
W : The set of all wicket keepers.
B : The set of all bowlers.

The above Venn diagram represents the truth of the given statement i.e., W ∩ B = φ.

EXERCISE 1.10Q 3.1  PAGE 27
Exercise 1.10 | Q 3.1 | Page 27

Represent the following statement by the Venn diagram.

Some non-resident Indians are not rich.

#### SOLUTION

Let, U : The set of all human beings.
N : The set of all non-resident Indians.
R : The set of all rich people.

The above Venn diagram represents the truth of the given statement i.e., N - R ≠ φ

EXERCISE 1.10Q 3.2  PAGE 27
Exercise 1.10 | Q 3.2 | Page 27

Represent the following statement by the Venn diagram.

No circle is rectangle.

#### SOLUTION

Let, U : The set of all geometrical figures.
C : The set of all circles.
R : The set of all rectangles.

The above Venn diagram represents the truth of the given statement i.e., C ∩ R = φ.

EXERCISE 1.10Q 3.3  PAGE 27
Exercise 1.10 | Q 3.3 | Page 27

Represent the following statement by the Venn diagram.

If n is a prime number and n ≠ 2, then it is odd.

#### SOLUTION

Let, U : The set of all real numbers.
P : The set of all prime numbers n and n ≠ 2.
O : The set of all odd numbers.

The above Venn diagram represents the truth of the given statement i.e., P ⊂ O.

## MISCELLANEOUS EXERCISE 1 [PAGES 29 - 34]

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Miscellaneous Exercise 1 [Pages 29 - 34]

#### QUESTION

Miscellaneous Exercise 1 | Q 1.01 | Page 29

Choose the correct alternative :

Which of the following is not a statement?

• Smoking is injuries to health

• 2 + 2 = 4

• 2 is the only even prime number.

• Come here

• Come here

#### QUESTION

Miscellaneous Exercise 1 | Q 1.02 | Page 29

Choose the correct alternative :

Which of the following is an open statement?

• x is a natural number.

• Give answer a glass of water.

• WIsh you best of luck.

• Good morning to all.

#### SOLUTION

x is a natural number.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.03 | Page 29

Choose the correct alternative :

Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as.

• commutative law

• associative law

• De-Morgan's law

• distributive law

#### SOLUTION

distributive law.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.04 | Page 29

Choose the correct alternative :

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

#### SOLUTION

(p → q) ↔ (∼ q → ∼ p) is a contradiction.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.05 | Page 29

Choose the correct alternative :

For the following three statements
p : 2 is an even number.
q : 2 is a prime number.
r : Sum of two prime numbers is always even.
Then, the symbolic statement (p ∧ q) → ∼ r means.

• 2 is an even and prime number and the sum of two prime numbers is always even.

• 2 is an even and prime number and the sum of two prime numbers is not always even.

• If 2 is an even and prime number, then the sum of two prime numbers is not always even.

• If 2 is an even and prime number, then the sum of two prime numbers is also even.

#### SOLUTION

If 2 is an even and prime number, then the sum of two prime numbers is not always even.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.06 | Page 30

Choose the correct alternative :

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)

#### QUESTION

Miscellaneous Exercise 1 | Q 1.07 | Page 30

Choose the correct alternative :

The negation of the proposition “If 2 is prime, then 3 is odd”, is

• If 2 is not prime, then 3 is not odd.

• 2 is prime and 3 is not odd.

• 2 is not prime and 3 is odd.

• If 2 is not prime, then 3 is odd.

#### SOLUTION

2 is prime and 3 is not odd.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.08 | Page 30

Choose the correct alternative :

The statement (∼ p ∧ q) ∨∼ q is

• p ∨ q

• p ∧ q

• ∼ (p ∨ q)

• ∼ (p ∧ q)

∼ (p ∧ q).

#### QUESTION

Miscellaneous Exercise 1 | Q 1.09 | Page 30

Choose the correct alternative :

Which of the following is always true?

• (p → q) ≡ ∼ q → ∼ p

• ∼ (p ∨ q) ≡ ∼ p ∨ ∼ q

• ∼ (p → q) ≡ p ∧ ∼ q

• ∼ (p ∨ q) ≡ ∼ p ∧ ∼ q

#### SOLUTION

∼ (p → q) ≡ p ∧ ∼ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.1 | Page 30

Choose the correct alternative :

∼ (p ∨ q) ∨ (∼ p ∧ q) is logically equivalent to

• ∼ p

• p

• q

• ∼ q

∼ p.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.11 | Page 30

Choose the correct alternative :

If p and q are two statements then (p → q) ↔ (∼ q → ∼ p) is

• tautology

• Neither (i) not (ii)

• None of the these

tautology.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.12 | Page 30

Choose the correct alternative :

If p is the sentence ‘This statement is false’ then

• truth value of p is T

• truth value of p is F

• p is both true and false

• p is neither true nor false

#### SOLUTION

p is neither true nor false

#### QUESTION

Miscellaneous Exercise 1 | Q 1.13 | Page 30

Choose the correct alternative :

Conditional p → q is equivalent to

• p → ∼ q

• ∼ p ∨ q

• ∼ p → ∼ q

• p ∨∼q

∼p ∨ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.14 | Page 30

Choose the correct alternative :

Negation of the statement “This is false or That is true” is

• That is true or This is false

• That is true and This is false

• That is true and That is false

• That is false and That is true

#### SOLUTION

That is true and That is false.

#### QUESTION

Miscellaneous Exercise 1 | Q 1.15 | Page 30

Choose the correct alternative :

If p is any statement then (p ∨ ∼ p) is a

• contingency

• tautology

• None of them

tautology.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.1 | Page 30

Fill in the blanks :

The statement q → p is called as the ––––––––– of the statement p → q.

#### SOLUTION

The statement q → p is called as the Converse of the statement p → q.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.2 | Page 30

Fill in the blanks :

Conjunction of two statement p and q is symbolically written as –––––––––.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.3 | Page 30

Fill in the blanks :

If p ∨ q is true then truth value of ∼ p ∨ ∼ q is –––––––––.

#### SOLUTION

If p ∨ q is true then truth value of ∼ p ∨ ∼ q is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.4 | Page 30

Fill in the blanks :

Negation of “some men are animal” is –––––––––.

#### SOLUTION

Negation of “some men are animal” is No men are animals.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.5 | Page 30

Fill in the blanks :

Truth value of if x = 2, then x2 = − 4 is –––––––––.

#### SOLUTION

Truth value of if x = 2, then x2 = − 4 is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.6 | Page 30

Fill in the blanks :

Inverse of statement pattern p ↔ q is given by –––––––––.

#### SOLUTION

Inverse of statement pattern p ↔ q is given by ∼ p → ∼ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.7 | Page 30

Fill in the blanks :

p ↔ q is false when p and q have ––––––––– truth values.

#### SOLUTION

p ↔ q is false when p and q have different truth values.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.8 | Page 31

Fill in the blanks :

Let p : the problem is easy. r : It is not challenging then verbal form of ∼ p → r is –––––––––.

#### SOLUTION

Let p : the problem is easy. r : It is not challenging then verbal form of ∼ p → r is If the problem is not easy them it is not challenging.

#### QUESTION

Miscellaneous Exercise 1 | Q 2.9 | Page 31

Fill in the blanks :

Truth value of 2 + 3 = 5 if and only if − 3 > − 9 is –––––––––.

#### SOLUTION

Truth value of 2 + 3 = 5 if and only if − 3 > − 9 is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.01 | Page 31

State whether the following statement is True or False :

Truth value of 2 + 3 < 6 is F.

• True

• False

False.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.02 | Page 31

State whether the following statement is True or False :

There are 24 months in year is a statement.

• True

• False

True.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.03 | Page 31

State whether the following statement is True or False :

p ∨ q has truth value F is both p and q has truth value F.

• True

• False

False.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.04 | Page 31

State whether the following statement is True or False :

The negation of 10 + 20 = 30 is, it is false that 10 + 20 ≠ 30.

• True

• False

False.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.05 | Page 31

State whether the following statement is True or False :

Dual of (p ∧ ∼ q) ∨ t is (p ∨ ∼ q) ∨ C.

• True

• False

False.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.06 | Page 31

State whether the following statement is True or False :

Dual of “John and Ayub went to the forest” is “John and Ayub went to the forest”.

• True

• False

True.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.07 | Page 31

State whether the following statement is True or False :

“His birthday is on 29th February” is not a statement.

• True

• False

True.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.08 | Page 31

State whether the following statement is True or False :

x2 = 25 is true statement.

• True

• False

False.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.09 | Page 31

State whether the following statement is True or False :

Truth value of $5$ √5 is not an irrational number is T.

• True

• False

False.

#### QUESTION

Miscellaneous Exercise 1 | Q 3.1 | Page 31

State whether the following statement is True or False :

p ∧ t = p.

• True

• False

True.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
Ice cream Sundaes are my favourite.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
x + 3 = 8 ; x is variable.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
Read a lot to improve your writing skill.

• Is a statement

• Is not a statement

#### SOLUTION

Is not a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
z is a positive number.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
(a + b)2 = a2 + 2ab + b2 for all a, b ∈ R.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
(2 + 1)2 = 9.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
Why are you sad?

• Is a statement

• Is not a statement

#### SOLUTION

Is not a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
How beautiful the flower is!

• Is a statement

• Is not a statement

#### SOLUTION

Is not a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
The square of any odd number is even.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
All integers are natural numbers.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
If x is real number then x2 ≥ 0.

• Is a statement

• Is not a statement

Is a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
Do not come inside the room.

• Is a statement

• Is not a statement

#### SOLUTION

Is not a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.
What a horrible sight it was!

• Is a statement

• Is not a statement

#### SOLUTION

Is not a statement

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

The square of every real number is positive.

• Is a statement

• Is not a statement

#### SOLUTION

It is a statement which is false. Hence, its truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

Every parallelogram is a rhombus.

• Is a statement

• Is not a statement

#### SOLUTION

It is a statement which is false. Hence, its truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

a2 − b2 = (a + b) (a − b) for all a, b ∈ R.

• Is a statement

• Is not a statement

#### SOLUTION

It is a statement which is true. Hence, its truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

Please carry out my instruction.

• Is a statement

• Is not a statement

#### SOLUTION

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

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

The Himalayas is the highest mountain range.

• Is a statement

• Is not a statement

#### SOLUTION

It is a statement which is true. Hence, its truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

(x − 2) (x − 3) = x2 − 5x + 6 for all x∈R.

• Is a statement

• Is not a statement

#### SOLUTION

It is a statement which is true. Hence, its truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

What are the causes of rural unemployment?

• Is a statement

• Is not a statement

#### SOLUTION

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

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

0! = 1

• Is a statement

• Is not a statement

#### SOLUTION

It is a statement which is true. Hence, its truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

The quadratic equation ax2 + bx + c = 0 (a ≠ 0) always has two real roots.

• Is a statement

• Is not a statement

#### SOLUTION

The quadratic equation ax2 + bx + c = 0 (a ≠ 0) always has two real roots is a statement.

Hence, its truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.02 | Page 31

Which of the following sentence is a statement? In case of a statement, write down the truth value.

What is happy ending?

• Is a statement

• Is not a statement

#### SOLUTION

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

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

The Sun has set and Moon has risen.

#### SOLUTION

Let p : The sun has set.
q : The moon has risen

The symbolic form is p ∧ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

Mona likes Mathematics and Physics.

#### SOLUTION

Let p : Mona likes Mathematics
q : Mona likes Physics

The symbolic form is p ∧ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

3 is prime number if 3 is perfect square number.

#### SOLUTION

Let p : 3 is a prime number.
q : 3 is a perfect square number.

The symbolic form is p ↔ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

Kavita is brilliant and brave.

#### SOLUTION

Let p : Kavita is brilliant.
q : Kavita is brave.

The symbolic form is p ∧ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

If Kiran drives the car, then Sameer will walk.

#### SOLUTION

Let p : Kiran drives the car.

q : Sameer will walk.

The symbolic form is p → q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

The necessary condition for existence of a tangent to the curve of the function is continuity.

#### SOLUTION

The given statement can also be expressed as ‘If the function is continuous, then the tangent to the curve exists’.

Let p : The function is continuous
q : The tangent to the curve exists.

∴ p → q is the symbolic form of the given statement.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

To be brave is necessary and sufficient condition to climb the Mount Everest.

#### SOLUTION

Let p : To be brave

q : climb the Mount Everest

∴ p ↔ q is the symbolic form of the given statement.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

x3 + y3 = (x + y)3 if xy = 0.

#### SOLUTION

Let p : x3 + y3 = (x + y)3

q : xy = 0

∴ p ↔ q is the symbolic form of the given statement.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 31

Assuming the first statement p and second as q. Write the following statement in symbolic form.

The drug is effective though it has side effects.

#### SOLUTION

The given statement can also be expressed as “The drug is effective and it has side effects”

Let p : The drug is effective.
q : It has side effects.

∴ p ∧ q is the symbolic form of the given statement.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 32

Assuming the first statement p and second as q. Write the following statement in symbolic form.

If a real number is not rational, then it must be irrational.

#### SOLUTION

Let p : A real number is not rational.
q : A real number must be irrational.

The symbolic form is p → q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 32

Assuming the first statement p and second as q. Write the following statement in symbolic form.

It is not true that Ram is tall and handsome.

#### SOLUTION

Let p : Ram is tall.
q : Ram is handsome.

The symbolic form is ∼(p ∧ q).

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 32

Assuming the first statement p and second as q. 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 still raining.

The symbolic form is ~ p ∧ q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 32

Assuming the first statement p and second as q. Write the following statement in symbolic form.

It is not true that intelligent persons are neither polite nor helpful.

#### SOLUTION

Let p : Intelligent persons are neither polite nor helpful

The symbolic form is ∼ p.

Alternate method:

Let p : Intelligent persons are polite.
q : Intelligent persons are helpful.
The symbolic form is ~(~ p ∧ ~ q).

#### QUESTION

Miscellaneous Exercise 1 | Q 4.03 | Page 32

Assuming the first statement p and second as q. Write the following statement in symbolic form.

If the question paper is not easy then we shall not pass.

#### SOLUTION

Let p : The question paper is not easy.
q : We shall not pass.

The symbolic form is p → q.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.04 | Page 32

If p : Proof is lengthy.
q : It is interesting.
Express the following statement in symbolic form.

Proof is lengthy and it is not interesting.

p ∧ ∼ q

#### QUESTION

Miscellaneous Exercise 1 | Q 4.04 | Page 32

If p : Proof is lengthy.
q : It is interesting.
Express the following statement in symbolic form.

If proof is lengthy then it is interesting.

p → q

#### QUESTION

Miscellaneous Exercise 1 | Q 4.04 | Page 32

If p : Proof is lengthy.
q : It is interesting.
Express the following statement in symbolic form.

It is not true that the proof is lengthy but it is interesting.

∼(p ∧ q)

#### QUESTION

Miscellaneous Exercise 1 | Q 4.04 | Page 32

If p : Proof is lengthy.
q : It is interesting.
Express the following statement in symbolic form.

It is interesting iff the proof is lengthy.

q ↔ p

#### QUESTION

Miscellaneous Exercise 1 | Q 4.05 | Page 32

Let p : Sachin wins the match.
q : Sachin is a member of Rajya Sabha.
r :
Sachin is happy.
Write the verbal statement of the following.

(p ∧ q) ∨ r

#### SOLUTION

Sachin wins the match or he is the member of Rajya Sabha or Sachin is happy.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.05 | Page 32

Let p : Sachin wins the match.
q : Sachin is a member of Rajya Sabha.
r : Sachin is happy.
Write the verbal statement of the following.

p → r

#### SOLUTION

If Sachin wins the match then he is happy.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.05 | Page 32

Let p : Sachin wins the match.
q : Sachin is a member of Rajya Sabha.
r : Sachin is happy.
Write the verbal statement of the following.

∼ p ∨ q

#### SOLUTION

Sachin does not win the match or he is the member of Rajya Sabha.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.05 | Page 32

Let p : Sachin wins the match.
q : Sachin is a member of Rajya Sabha.
r : Sachin is happy.
Write the verbal statement of the following.

p → (p ∧ r)

#### SOLUTION

If sachin wins the match, then he is the member of Rajyasabha or he is happy.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.05 | Page 32

Let p : Sachin wins the match.
q : Sachin is a member of Rajya Sabha.
r : Sachin is happy.
Write the verbal statement of the following.

p → q

#### SOLUTION

If Sachin wins the match then he is a member of Rajyasabha.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.05 | Page 32

Let p : Sachin wins the match.
q : Sachin is a member of Rajya Sabha.
r : Sachin is happy.
Write the verbal statement of the following.

(p ∧ q) ∧ ∼ r

#### SOLUTION

Sachin wins the match and he is the member of Rajyasabha but he is not happy.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.05 | Page 32

Let p : Sachin wins the match.
q : Sachin is a member of Rajya Sabha.
r : Sachin is happy.
Write the verbal statement of the following.

∼ (p ∨ q) ∧ r

#### SOLUTION

It is false that Sachin wins the match or he is the member of Rajyasabha but he is happy.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.06 | Page 32

Determine the truth value of the following statement.

4 + 5 = 7 or 9 − 2 = 5

#### SOLUTION

Let p : 4 + 5 = 7
q : 9 – 2 = 5
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.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.06 | Page 32

Determine the truth value of the following statement.

If 9 > 1 then x2 − 2x + 1 = 0 for x = 1

#### SOLUTION

Let p : 9 > 1
q : x2 – 2x + 1 = 0 for x = 1

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.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.06 | Page 32

Determine the truth value of the following statement.

x + y = 0 is the equation of a straight line if and only if y2 = 4x is the equation of the parabola.

#### SOLUTION

Let p : x + y = 0 is the equation of a straight line.
q : y2 = 4x is the equation of the parabola.

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.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.06 | Page 32

Determine the truth value of the following statement.

It is not true that 2 + 3 = 6 or 12 + 3 =5

#### SOLUTION

Let p : 2 + 3 = 6
q : 12 + 3 = 5
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 ≡ T

∴ Truth value of the given statement is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.07 | Page 32

Assuming the following statement.
p : Stock prices are high.
q : Stocks are rising.
to be true, find the truth value of the following.

Stock prices are not high or stocks are rising.

#### SOLUTION

Given that the truth values of both p and q are T.

The symbolic form of the given statement is ~ p ∨ q.

∴ ~ p ∨ q ≡ ~ T ∨ T ≡ F ∨ T

Hence, truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.07 | Page 32

Assuming the following statement.
p : Stock prices are high.
q : Stocks are rising.
to be true, find the truth value of the following.

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

#### SOLUTION

The symbolic form of the given statement is
(p ∧ q) ↔ p.

∴ (p ∧ q) ↔ p ≡ (T ∧ T) ↔ T

≡ T ↔ T

≡ T

Hence, truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.07 | Page 32

Assuming the following statement.
p : Stock prices are high.
q : Stocks are rising.
to be true, find the truth value of the following.

If stock prices are high then stocks are not rising.

#### SOLUTION

The Symbolic form of the given statement is p → ~ q.

∴ p → ~ q ≡ T → ~ T ≡ T → F ≡ F

Hence, truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.07 | Page 32

Assuming the following statement.
p : Stock prices are high.
q : Stocks are rising.
to be true, find the truth value of the following.

It is false that stocks are rising and stock prices are high.

#### SOLUTION

The symbolic form of the given statement is ~(q ∧ p).

∴ ~(q ∧ p) ≡ ~(T ∧ T) ≡ ~T ≡ F

Hence, truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.07 | Page 32

Assuming the following statement.

p : Stock prices are high.

q : Stocks are rising.

to be true, find the truth value of the following.

Stock prices are high or stocks are not rising iff stocks are rising.

#### SOLUTION

The symbolic form of the given statement is (p ∨ ~q) ↔ q.

∴ (p ∨ ~q) ↔ q ≡ (T ∨ ~T) ↔ T

≡ (T ∨ F) ↔ T

≡ T ↔ T

≡ T

Hence, truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.08 | Page 32

Rewrite the following statement without using conditional –
(Hint : p → q ≡ ∼ p ∨ q)

If price increases, then demand falls.

#### SOLUTION

Let p : Prince increases.
q : demand falls.
The given statement is p → q.
But p → q ≡ ~p ∨ q.

The given statement can be written as ‘Price does not increase or demand falls’.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.08 | Page 32

Rewrite the following statement without using conditional –
(Hint : p → q ≡ ∼ p ∨ q)

If demand falls, then price does not increase.

#### SOLUTION

Let p : demand falls.
q : Price does not increase.
The given statement is p → q.

But p → q ≡ ~ p ∨ q.

∴ The given statement can be written as ‘Demand does not fall or price does not increase’.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.09 | Page 32

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

(p ∧ q) → ∼ p.

#### SOLUTION

(p ∧ q) → ∼ p ≡ (T ∧ T) → ∼ T

≡ T  → F

≡ F.

Hence, truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.09 | Page 32

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

p ↔ (q → ∼ p)

#### SOLUTION

p ↔ (q → ∼ p) ≡ T ↔ (T → ∼ T)

≡ T ↔ (T → F)

≡ T ↔ F

≡ F

Hence, truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.09 | Page 32

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

(p ∧ ∼ q) ∨ (∼ p ∧ q)

#### SOLUTION

(p ∧ ∼ q) ∨ (∼ p ∧ q) ≡ (T ∧ ∼ T) ∨ (∼ T ∧ T)

≡ (T ∧ F) ∨ (F ∧ T)

≡ F ∨ F

≡ F

Hence, truth value is F.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.09 | Page 32

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

∼ (p ∧ q) → ∼ (q ∧ p)

#### SOLUTION

∼ (p ∧ q) → ∼ (q ∧ p) ≡ ∼ (T ∧ T) → ∼ (T ∧ T)

≡ ~ T → ~ T

≡ F → F

≡ T

Hence, truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.09 | Page 32

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

∼ [(p → q) ↔ (p ∧ ∼ q)]

#### SOLUTION

∼[(p → q) ↔ (p ∧ ∼q)] ≡ ∼ [(T → T) ↔ (T ∧ ∼ T)]

≡ ~[T ↔ (T ∧ F)]

≡ ~(T ↔ F)

≡ ~ F

≡ T

Hence, truth value is T.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.1 | Page 32

Write the negation of the following.

If ∆ABC is not equilateral, then it is not equiangular.

#### SOLUTION

Let p : ∆ ABC is not equilateral.
q : ∆ ABC is not equiangular.

The given statement is p → q.

Its negation is ~(p → q) ≡ p ∧ ~ q

∴ The negation of given statement is '∆ ABC is not equilateral and it is equiangular'.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.1 | Page 32

Write the negation of the following.

Ramesh is intelligent and he is hard working.

#### SOLUTION

Let p : Ramesh is intelligent.
q : Ramesh is hard working.
The given statement is p ∧ q.

Its negation is ~(p ∧ q) ≡ ~ p ∨ ~ q

∴ The negation of the given statement is ‘Ramesh is not intelligent or he is not hard-working.’

#### QUESTION

Miscellaneous Exercise 1 | Q 4.1 | Page 32

Write the negation of the following.

An angle is a right angle if and only if it is of measure 90°.

#### SOLUTION

Let p : An angle is a right angle.
q : An angle is of measure 90°.

The given statement is p ↔ q.

Its negation is ~(p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p)

∴ The negation of the given statement is ‘An angle is a right angle and it is not of measure 90° or an angle is of measure 90° and it is not a right angle.’

#### QUESTION

Miscellaneous Exercise 1 | Q 4.1 | Page 32

Write the negation of the following.

Kanchanganga is in India and Everest is in Nepal.

#### SOLUTION

Let p : Kanchanganga is in India.
q : Everest is in Nepal.
The given statement is p ∧ q.

Its negation is ~(p ∧ q) ≡ ~ p ∨ ~ q.

The negation of a given statement is ‘Kanchanganga is not in India or Everest is not in Nepal’.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.1 | Page 32

Write the negation of the following.

If x ∈ A ∩ B, then x ∈ A and x ∈ B.

#### SOLUTION

Let p : x ∈ A ∩ B

q : x ∈ A

r : x ∈ B

The given statement is p → (q ∧ r).

Its negation is ~[p → (q ∧ r)], and

~[p → (q ∧ r)] ≡ p ∧ ~ (q ∧ r) ≡ p ∧ ~ q ∨ ~ r

∴ The negation of given statement is x ∈ A ∩ B and x ∉ A or x ∉ B.

#### QUESTION

Miscellaneous Exercise 1 | Q 4.11 | Page 33

Construct the truth table for the following statement pattern.

(p ∧ ~q) ↔ (q → p)

#### SOLUTION

 p q ~q p∧~q q→p (p∧~q)↔(q→p) T T F F T F T F T T T T F T F F F T F F T F T F