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

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 injurious to health
  • \(2 + 2 = 4\)
  • 2 is the only even prime number.
  • Come here
SOLUTION

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 me 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 \(p \land (q \lor r) \equiv (p \land q) \lor (p \land r)\). 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

  • \(p \land (\sim p)\) is contradiction
  • \((p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)\) is a contradiction.
  • \(\sim (\sim p) \leftrightarrow p\) is a tautology
  • \(p \lor (\sim p)\) is a tautology
SOLUTION

\((p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)\) 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 \((p \land q) \rightarrow \sim r\) 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 :

\(p\) : He is intelligent.
\(q\) : He is strong
Then, symbolic form of statement “It is wrong that, he is intelligent or strong” is

  • \(\sim p \lor \sim q\)
  • \(\sim (p \land q)\)
  • \(\sim (p \lor q)\)
  • \(p \lor \sim q\)
SOLUTION

\(\sim (p \lor q)\)

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 \((\sim p \land q) \lor \sim q\) is

  • \(p \lor q\)
  • \(p \land q\)
  • \(\sim (p \lor q)\)
  • \(\sim (p \land q)\)
SOLUTION

\(\sim (p \land q)\).

QUESTION Miscellaneous Exercise 1 | Q 1.09 | Page 30

Choose the correct alternative :

Which of the following is always true?

  • \((p \rightarrow q) \equiv \sim q \rightarrow \sim p\)
  • \(\sim (p \lor q) \equiv \sim p \lor \sim q\)
  • \(\sim (p \rightarrow q) \equiv p \land \sim q\)
  • \(\sim (p \land q) \equiv \sim p \land \sim q\)
SOLUTION

\((p \rightarrow q) \equiv \sim q \rightarrow \sim p\)

QUESTION Miscellaneous Exercise 1 | Q 1.10 | Page 30

Choose the correct alternative :

\(\sim (p \lor q) \lor (\sim p \land q)\) is logically equivalent to

  • \(\sim p\)
  • \(p\)
  • \(q\)
  • \(\sim q\)
SOLUTION

\(\sim p\).

QUESTION Miscellaneous Exercise 1 | Q 1.11 | Page 30

Choose the correct alternative :

If \(p\) and \(q\) are two statements then \((p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)\) is

  • contradiction
  • tautology
  • Neither (i) nor (ii)
  • None of these
SOLUTION

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 \(p \rightarrow q\) is equivalent to

  • \(p \rightarrow \sim q\)
  • \(\sim p \lor q\)
  • \(\sim p \rightarrow \sim q\)
  • \(p \lor \sim q\)
SOLUTION

\(\sim p \lor q\).

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
  • This is true and That is false
  • That is true and That is false
  • That is false and That is true
SOLUTION

This 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 \((p \lor \sim p)\) is a

  • contingency
  • contradiction
  • tautology
  • None of them
SOLUTION

tautology.

QUESTION Miscellaneous Exercise 1 | Q 2.1 | Page 30

Fill in the blanks :

The statement \(q \rightarrow p\) is called as the ––––––––– of the statement \(p \rightarrow q\).

SOLUTION

The statement \(q \rightarrow p\) is called as the Converse of the statement \(p \rightarrow q\).

QUESTION Miscellaneous Exercise 1 | Q 2.2 | Page 30

Fill in the blanks :

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

SOLUTION

Conjunction of two statement \(p\) and \(q\) is symbolically written as \(p \land q\).

QUESTION Miscellaneous Exercise 1 | Q 2.3 | Page 30

Fill in the blanks :

If \(p \lor q\) is true then truth value of \(\sim p \lor \sim q\) is –––––––––.

SOLUTION

If \(p \lor q\) is true then truth value of \(\sim p \lor \sim q\) is F. (This holds if \(p=T, q=T\). The question is ambiguous for other cases where \(p \lor q\) is true.)

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. (Or "All men are not animals.")

QUESTION Miscellaneous Exercise 1 | Q 2.5 | Page 30

Fill in the blanks :

Truth value of if \(x = 2\), then \(x^2 = -4\) is –––––––––.

SOLUTION

Truth value of if \(x = 2\), then \(x^2 = -4\) is F.

QUESTION Miscellaneous Exercise 1 | Q 2.6 | Page 30

Fill in the blanks :

Inverse of statement pattern \(p \leftrightarrow q\) is given by –––––––––.

SOLUTION

Inverse of statement pattern \(p \leftrightarrow q\) is given by \(\sim p \leftrightarrow \sim q\).

QUESTION Miscellaneous Exercise 1 | Q 2.7 | Page 30

Fill in the blanks :

\(p \leftrightarrow q\) is false when \(p\) and \(q\) have ––––––––– truth values.

SOLUTION

\(p \leftrightarrow q\) 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 \(\sim p \rightarrow r\) is –––––––––.

SOLUTION

Let \(p\) : the problem is easy. \(r\) : It is not challenging then verbal form of \(\sim p \rightarrow r\) is If the problem is not easy then 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.

SOLUTION

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.

SOLUTION

True.

QUESTION Miscellaneous Exercise 1 | Q 3.03 | Page 31

State whether the following statement is True or False :

\(p \lor q\) has truth value F if both \(p\) and \(q\) have truth value F.

SOLUTION

True.

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 \(10 + 20 \neq 30\).

SOLUTION

True.

QUESTION Miscellaneous Exercise 1 | Q 3.05 | Page 31

State whether the following statement is True or False :

Dual of \((p \land \sim q) \lor t\) is \((p \lor \sim q) \lor C\).

SOLUTION

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”.

SOLUTION

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.

SOLUTION

True.

QUESTION Miscellaneous Exercise 1 | Q 3.08 | Page 31

State whether the following statement is True or False :

\(x^2 = 25\) is true statement.

SOLUTION

False.

QUESTION Miscellaneous Exercise 1 | Q 3.09 | Page 31

State whether the following statement is True or False :

Truth value of \(\sqrt{5}\) is not an irrational number is T.

SOLUTION

False.

QUESTION Miscellaneous Exercise 1 | Q 3.10 | Page 31

State whether the following statement is True or False :

\(p \land t \equiv p\).

SOLUTION

True.

QUESTION Miscellaneous Exercise 1 | Q 4.01 | Page 31

Solve the following :

State which of the following sentences are statements in logic.

  1. Ice cream Sundaes are my favourite.
  2. \(x + 3 = 8\) ; \(x\) is variable.
  3. Read a lot to improve your writing skill.
  4. \(z\) is a positive number.
  5. \((a + b)^2 = a^2 + 2ab + b^2\) for all \(a, b \in R\).
  6. \((2 + 1)^2 = 9\).
  7. Why are you sad?
  8. How beautiful the flower is!
  9. The square of any odd number is even.
  10. All integers are natural numbers.
  11. If \(x\) is real number then \(x^2 \ge 0\).
  12. Do not come inside the room.
  13. What a horrible sight it was!
SOLUTION
  1. Is not a statement (Subjective opinion).
  2. Is not a statement (Open sentence).
  3. Is not a statement (Imperative sentence/advice).
  4. Is not a statement (Open sentence).
  5. Is a statement (It is a mathematical identity, which is true).
  6. Is a statement (It is \(3^2=9\), which is \(9=9\), a true statement).
  7. Is not a statement (Interrogative sentence).
  8. Is not a statement (Exclamatory sentence).
  9. Is a statement (It is a declarative sentence which is false).
  10. Is a statement (It is a declarative sentence which is false).
  11. Is a statement (It is a declarative sentence which is true).
  12. Is not a statement (Imperative sentence/command).
  13. Is not a statement (Exclamatory sentence).
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.

  1. The square of every real number is positive.
  2. Every parallelogram is a rhombus.
  3. \(a^2 - b^2 = (a + b) (a - b)\) for all \(a, b \in R\).
  4. Please carry out my instruction.
  5. The Himalayas is the highest mountain range.
  6. \((x - 2) (x - 3) = x^2 - 5x + 6\) for all \(x \in R\).
  7. What are the causes of rural unemployment?
  8. \(0! = 1\)
  9. The quadratic equation \(ax^2 + bx + c = 0\) (\(a \neq 0\)) always has two real roots.
  10. What is happy ending?
SOLUTION
  1. It is a statement. The square of \(0\) (a real number) is \(0\), which is not positive. Hence, its truth value is F.
  2. It is a statement. Not every parallelogram is a rhombus (e.g., a rectangle that is not a square). Hence, its truth value is F.
  3. It is a statement. This is a standard algebraic identity. Hence, its truth value is T.
  4. It is an imperative sentence (a request). Hence, it is not a statement.
  5. It is a statement. The Himalayas are the highest mountain range in the world. Hence, its truth value is T.
  6. It is a statement. This is a standard algebraic identity. Hence, its truth value is T.
  7. It is an interrogative sentence. Hence, it’s not a statement.
  8. It is a statement. By definition, \(0! = 1\). Hence, its truth value is T.
  9. It is a statement. A quadratic equation has real roots if its discriminant \(b^2 - 4ac \ge 0\). It does not always have two real roots (e.g., if \(b^2 - 4ac < 0\), it has complex roots). Hence, its truth value is F.
  10. It is an interrogative sentence. Hence, it’s not a statement.
QUESTION Miscellaneous Exercise 1 | Q 4.03 | Page 31-32

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

  1. The Sun has set and Moon has risen.
  2. Mona likes Mathematics and Physics.
  3. 3 is prime number if and only if 3 is perfect square number. (Corrected "if" to "if and only if" based on solution)
  4. Kavita is brilliant and brave.
  5. If Kiran drives the car, then Sameer will walk.
  6. The necessary condition for existence of a tangent to the curve of the function is continuity.
  7. To be brave is necessary and sufficient condition to climb the Mount Everest.
  8. \(x^3 + y^3 = (x + y)^3\) if and only if \(xy = 0\). (Corrected "if" to "if and only if" based on solution)
  9. The drug is effective though it has side effects.
  10. If a real number is not rational, then it must be irrational.
  11. It is not true that Ram is tall and handsome.
  12. Even though it is not cloudy, it is still raining.
  13. It is not true that intelligent persons are neither polite nor helpful.
  14. If the question paper is not easy then we shall not pass.
SOLUTION
  1. Let \(p\): The sun has set. \(q\): The moon has risen. Symbolic form: \(p \land q\).
  2. Let \(p\): Mona likes Mathematics. \(q\): Mona likes Physics. Symbolic form: \(p \land q\).
  3. Let \(p\): 3 is a prime number. \(q\): 3 is a perfect square number. Symbolic form: \(p \leftrightarrow q\).
  4. Let \(p\): Kavita is brilliant. \(q\): Kavita is brave. Symbolic form: \(p \land q\).
  5. Let \(p\): Kiran drives the car. \(q\): Sameer will walk. Symbolic form: \(p \rightarrow q\).
  6. Let \(p\): Existence of a tangent to the curve of the function. \(q\): The function is continuous. "q is necessary for p" means "if p then q" (\(p \rightarrow q\)). Or, "If the tangent exists, then the function is continuous." The original solution was "If the function is continuous, then the tangent to the curve exists." Let's use this interpretation. \(p\): The function is continuous. \(q\): The tangent to the curve exists. Symbolic form: \(p \rightarrow q\).
  7. Let \(p\): One is brave. \(q\): One can climb Mount Everest. Symbolic form: \(p \leftrightarrow q\).
  8. Let \(p\): \(x^3 + y^3 = (x + y)^3\). \(q\): \(xy = 0\). Symbolic form: \(p \leftrightarrow q\).
  9. "Though" implies "and". Let \(p\): The drug is effective. \(q\): It has side effects. Symbolic form: \(p \land q\).
  10. Let \(p\): A real number is not rational. \(q\): It must be irrational. Symbolic form: \(p \rightarrow q\).
  11. Let \(p\): Ram is tall. \(q\): Ram is handsome. Symbolic form: \(\sim(p \land q)\).
  12. Let \(p\): It is cloudy. \(q\): It is still raining. Symbolic form: \(\sim p \land q\).
  13. Let \(p'\): Intelligent persons are polite. \(q'\): Intelligent persons are helpful. The statement "intelligent persons are neither polite nor helpful" is \(\sim p' \land \sim q'\). "It is not true that..." means negation of this. Symbolic form: \(\sim (\sim p' \land \sim q')\). This is equivalent to \(p' \lor q'\). (Intelligent persons are polite or helpful). Alternatively, if \(r\): Intelligent persons are neither polite nor helpful. Symbolic form: \(\sim r\).
  14. Let \(p\): The question paper is not easy. \(q\): We shall not pass. Symbolic form: \(p \rightarrow q\).
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.

  1. Proof is lengthy and it is not interesting.
  2. If proof is lengthy then it is interesting.
  3. It is not true that the proof is lengthy but it is interesting.
  4. It is interesting iff the proof is lengthy.
SOLUTION
  1. \(p \land \sim q\)
  2. \(p \rightarrow q\)
  3. \(\sim (p \land q)\)
  4. \(q \leftrightarrow p\)
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.

  1. \((p \land q) \lor r\)
  2. \(p \rightarrow r\)
  3. \(\sim p \lor q\)
  4. \(p \rightarrow (q \lor r)\)
  5. \(p \rightarrow q\)
  6. \((p \land q) \land \sim r\)
  7. \(\sim (p \lor q) \land r\)
SOLUTION
  1. Sachin wins the match and he is a member of Rajya Sabha, or Sachin is happy.
  2. If Sachin wins the match then he is happy.
  3. Sachin does not win the match or he is a member of Rajya Sabha.
  4. If Sachin wins the match, then he is a member of Rajya Sabha or he is happy.
  5. If Sachin wins the match then he is a member of Rajya Sabha.
  6. Sachin wins the match and he is a member of Rajya Sabha, but he is not happy.
  7. It is not true that Sachin wins the match or is a member of Rajya Sabha, and Sachin is happy. (Alternatively: Neither Sachin wins the match nor is he a member of Rajya Sabha, but he is happy.)
QUESTION Miscellaneous Exercise 1 | Q 4.06 | Page 32

Determine the truth value of the following statement.

  1. \(4 + 5 = 7\) or \(9 - 2 = 5\)
  2. If \(9 > 1\) then \(x^2 - 2x + 1 = 0\) for \(x = 1\)
  3. \(x + y = 0\) is the equation of a straight line if and only if \(y^2 = 4x\) is the equation of the parabola.
  4. It is not true that \(2 + 3 = 6\) or \(12 + 3 = 5\)
SOLUTION
  1. Let \(p: 4 + 5 = 7\) (False). Let \(q: 9 - 2 = 5\) (False). The statement is \(p \lor q\). \(F \lor F \equiv F\). Truth value is F.
  2. Let \(p: 9 > 1\) (True). Let \(q: x^2 - 2x + 1 = 0\) for \(x = 1\). For \(x=1\), \(1^2 - 2(1) + 1 = 1 - 2 + 1 = 0\), so \(q\) is True. The statement is \(p \rightarrow q\). \(T \rightarrow T \equiv T\). Truth value is T.
  3. Let \(p: x + y = 0\) is the equation of a straight line (True). Let \(q: y^2 = 4x\) is the equation of a parabola (True). The statement is \(p \leftrightarrow q\). \(T \leftrightarrow T \equiv T\). Truth value is T.
  4. Let \(p: 2 + 3 = 6\) (False). Let \(q: 12 + 3 = 5\) (False). The statement is \(\sim (p \lor q)\). \(\sim (F \lor F) \equiv \sim F \equiv T\). 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.

  1. Stock prices are not high or stocks are rising.
  2. Stock prices are high and stocks are rising if and only if stock prices are high.
  3. If stock prices are high then stocks are not rising.
  4. It is false that stocks are rising and stock prices are high.
  5. Stock prices are high or stocks are not rising iff stocks are rising.
SOLUTION

Given \(p\) is T, \(q\) is T.

  1. Symbolic form: \(\sim p \lor q\). Truth value: \(\sim T \lor T \equiv F \lor T \equiv T\).
  2. Symbolic form: \((p \land q) \leftrightarrow p\). Truth value: \((T \land T) \leftrightarrow T \equiv T \leftrightarrow T \equiv T\).
  3. Symbolic form: \(p \rightarrow \sim q\). Truth value: \(T \rightarrow \sim T \equiv T \rightarrow F \equiv F\).
  4. Symbolic form: \(\sim (q \land p)\). Truth value: \(\sim (T \land T) \equiv \sim T \equiv F\).
  5. Symbolic form: \((p \lor \sim q) \leftrightarrow q\). Truth value: \((T \lor \sim T) \leftrightarrow T \equiv (T \lor F) \leftrightarrow T \equiv T \leftrightarrow T \equiv T\).
QUESTION Miscellaneous Exercise 1 | Q 4.08 | Page 32

Rewrite the following statement without using conditional – (Hint : \(p \rightarrow q \equiv \sim p \lor q\))

  1. If price increases, then demand falls.
  2. If demand falls, then price does not increase.
SOLUTION
  1. Let \(p\): Price increases. \(q\): Demand falls. The statement is \(p \rightarrow q\), which is \(\sim p \lor q\). "Price does not increase or demand falls."
  2. Let \(p\): Demand falls. \(q\): Price does not increase. The statement is \(p \rightarrow q\), which is \(\sim p \lor q\). "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.

  1. \( (p \land q) \rightarrow \sim p \)
  2. \( p \leftrightarrow (q \rightarrow \sim r) \)
  3. \( (p \land \sim q) \lor (\sim p \land q) \)
  4. \( \sim (p \land q) \rightarrow \sim (q \land p) \)
  5. \( \sim [(p \rightarrow q) \leftrightarrow (p \land \sim q)] \)
SOLUTION

Given \(p \equiv T\), \(q \equiv T\), \(r \equiv F\).

  1. \( (T \land T) \rightarrow \sim T \equiv T \rightarrow F \equiv F \). Truth value is F.
  2. \( T \leftrightarrow (T \rightarrow \sim F) \equiv T \leftrightarrow (T \rightarrow T) \equiv T \leftrightarrow T \equiv T \). Truth value is T.
  3. \( (T \land \sim T) \lor (\sim T \land T) \equiv (T \land F) \lor (F \land T) \equiv F \lor F \equiv F \). Truth value is F.
  4. \( \sim (T \land T) \rightarrow \sim (T \land T) \equiv \sim T \rightarrow \sim T \equiv F \rightarrow F \equiv T \). Truth value is T.
  5. \( \sim [(T \rightarrow T) \leftrightarrow (T \land \sim T)] \equiv \sim [T \leftrightarrow (T \land F)] \equiv \sim [T \leftrightarrow F] \equiv \sim F \equiv T \). Truth value is T.
QUESTION Miscellaneous Exercise 1 | Q 4.10 | Page 32

Write the negation of the following.

  1. If \(\triangle ABC\) is not equilateral, then it is not equiangular.
  2. Ramesh is intelligent and he is hard working.
  3. An angle is a right angle if and only if it is of measure \(90^\circ\).
  4. Kanchenjunga is in India and Everest is in Nepal.
  5. If \(x \in A \cap B\), then \(x \in A\) and \(x \in B\).
SOLUTION
  1. Let \(p\): \(\triangle ABC\) is not equilateral. \(q\): It is not equiangular. Statement: \(p \rightarrow q\). Negation: \(\sim(p \rightarrow q) \equiv p \land \sim q\). Negation: "\(\triangle ABC\) is not equilateral and it is equiangular."
  2. Let \(p\): Ramesh is intelligent. \(q\): He is hard working. Statement: \(p \land q\). Negation: \(\sim(p \land q) \equiv \sim p \lor \sim q\). Negation: "Ramesh is not intelligent or he is not hard working."
  3. Let \(p\): An angle is a right angle. \(q\): It is of measure \(90^\circ\). Statement: \(p \leftrightarrow q\). Negation: \(\sim(p \leftrightarrow q) \equiv (p \land \sim q) \lor (q \land \sim p)\). Negation: "An angle is a right angle and it is not of measure \(90^\circ\), or an angle is of measure \(90^\circ\) and it is not a right angle."
  4. Let \(p\): Kanchenjunga is in India. \(q\): Everest is in Nepal. Statement: \(p \land q\). Negation: \(\sim(p \land q) \equiv \sim p \lor \sim q\). Negation: "Kanchenjunga is not in India or Everest is not in Nepal."
  5. Let \(p\): \(x \in A \cap B\). \(q\): \(x \in A\). \(r\): \(x \in B\). Statement: \(p \rightarrow (q \land r)\). Negation: \(\sim(p \rightarrow (q \land r)) \equiv p \land \sim(q \land r) \equiv p \land (\sim q \lor \sim r)\). Negation: "\(x \in A \cap B\), and (\(x \notin A\) or \(x \notin B\))."
QUESTION Miscellaneous Exercise 1 | Q 4.11 | Page 33

Construct the truth table for the following statement pattern.

  1. \( (p \land \sim q) \leftrightarrow (q \rightarrow p) \)
  2. \( (\sim p \lor q) \land (\sim p \land \sim q) \)
  3. \( (p \land r) \rightarrow (p \lor \sim q) \)
  4. \( (p \lor r) \rightarrow \sim (q \land r) \)
  5. \( (p \lor \sim q) \rightarrow (r \land p) \)
SOLUTION
  1. \( (p \land \sim q) \leftrightarrow (q \rightarrow p) \)
    pq\(\sim q\)\(p \land \sim q\)\(q \rightarrow p\)\((p \land \sim q) \leftrightarrow (q \rightarrow p)\)
    TTFFTF
    TFTTTT
    FTFFFT
    FFTFTF
  2. \( (\sim p \lor q) \land (\sim p \land \sim q) \)
    pq\(\sim p\)\(\sim q\)\(\sim p \lor q\)\(\sim p \land \sim q\)\((\sim p \lor q) \land (\sim p \land \sim q)\)
    TTFFTFF
    TFFTFFF
    FTTFTFF
    FFTTTTT
  3. \( (p \land r) \rightarrow (p \lor \sim q) \)
    pqr\(\sim q\)\(p \land r\)\(p \lor \sim q\)\((p \land r) \rightarrow (p \lor \sim q)\)
    TTTFTTT
    TTFFFTT
    TFTTTTT
    TFFTFTT
    FTTFFFT
    FTFFFFT
    FFTTFTT
    FFFTFTT
  4. \( (p \lor r) \rightarrow \sim (q \land r) \)
    pqr\(p \lor r\)\(q \land r\)\(\sim(q \land r)\)\((p \lor r) \rightarrow \sim(q \land r)\)
    TTTTTFF
    TTFTFTT
    TFTTFTT
    TFFTFTT
    FTTTTFF
    FTFFFTT
    FFTTFTT
    FFFFFTT
  5. \( (p \lor \sim q) \rightarrow (r \land p) \)
    pqr\(\sim q\)\(p \lor \sim q\)\(r \land p\)\((p \lor \sim q) \rightarrow (r \land p)\)
    TTTFTTT
    TTFFTFF
    TFTTTTT
    TFFTTFF
    FTTFFFT
    FTFFFFT
    FFTTTFF
    FFFTTFF
QUESTION Miscellaneous Exercise 1 | Q 4.12 | Page 33

What is tautology? What is contradiction?
Show that the negation of a tautology is a contradiction and the negation of a contradiction is a tautology.

SOLUTION

Tautology: A statement pattern having truth value always T, irrespective of the truth values of its component statement is called a tautology.

Contradiction: A statement pattern having truth value always F, irrespective of the truth values of its component statement is called a contradiction.

Let \(S_t\) be a tautology. This means \(S_t\) always has the truth value T.

\(S_t\)\(\sim S_t\)
TF

Since \(\sim S_t\) always has the truth value F, the negation of a tautology is a contradiction.

Let \(S_c\) be a contradiction. This means \(S_c\) always has the truth value F.

\(S_c\)\(\sim S_c\)
FT

Since \(\sim S_c\) always has the truth value T, the negation of a contradiction is a tautology.

QUESTION Miscellaneous Exercise 1 | Q 4.13 | Page 33

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

  1. \( [(p \land q) \lor (\sim p)] \lor [p \land (\sim q)] \)
  2. \( [(\sim p \land q) \land (q \land r)] \lor (\sim q) \)
  3. \( [(\sim p \lor q) \rightarrow p] \leftrightarrow [(\sim p) \land (\sim q)] \)
  4. \( [p \rightarrow (\sim q \lor r)] \leftrightarrow \sim [p \rightarrow (q \rightarrow r)] \)
SOLUTION
  1. \( [(p \land q) \lor (\sim p)] \lor [p \land (\sim q)] \)
    pq\(\sim p\)\(\sim q\)\(p \land q\)\((p \land q) \lor (\sim p)\)\(p \land \sim q\)LHS
    TTFFTTFT
    TFFTFFTT
    FTTFFTFT
    FFTTFTFT

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

  2. \( [(\sim p \land q) \land (q \land r)] \lor (\sim q) \)
    pqr\(\sim p\)\(\sim q\)\(\sim p \land q\)\(q \land r\)\((\sim p \land q) \land (q \land r)\)LHS
    TTTFFFTFF
    TTFFFFFFF
    TFTFTFFFT
    TFFFTFFFT
    FTTTFTTTT
    FTFTFTFFF
    FFTTTFFFT
    FFFTTFFFT

    Truth values in the last column (LHS) are not all T and not all F. Hence, it is a contingency.

  3. \( [(\sim p \lor q) \rightarrow p] \leftrightarrow [(\sim p) \land (\sim q)] \)
    pq\(\sim p\)\(\sim q\)\(\sim p \lor q\)\((\sim p \lor q) \rightarrow p\)\((\sim p) \land (\sim q)\)LHS \(\leftrightarrow\) RHS
    TTFFTTFF
    TFFTFTFF
    FTTFTFFT
    FFTTTFTF

    Truth values in the last column are not all T and not all F. Hence, it is a contingency. (Note: Original problem statement had \(\sim(p \lor q)\) as antecedent of first conditional, this is solved per original HTML. If the original was \(\sim(p \lor q) \rightarrow p\), the table would differ and result might change. I solved the formulation given in the original HTML for 4.13.iii)
    Re-checking the transcribed original: `[~(p ∨ q) → p] ↔ [(~p) ∧ (~q)]`. Let's do this one to match the original HTML structure more closely.

    pq\(\sim p\)\(\sim q\)\(p \lor q\)\(\sim(p \lor q)\)\(\sim(p \lor q) \rightarrow p\)\(\sim p \land \sim q\)LHS \(\leftrightarrow\) RHS
    TTFFTFTFF
    TFFTTFTFF
    FTTFTFTFF
    FFTTFTFTF

    All truth values in the last column are F. Hence, it is a contradiction. (This matches the original HTML for Q4.13.iii's solution which states this specific structure results in a contradiction.)

  4. \( [p \rightarrow (\sim q \lor r)] \leftrightarrow \sim [p \rightarrow (q \rightarrow r)] \)
    pqr\(\sim q\)\(\sim q \lor r\)\(p \rightarrow (\sim q \lor r)\) (A)\(q \rightarrow r\)\(p \rightarrow (q \rightarrow r)\) (B)\(\sim B\)A \(\leftrightarrow \sim B\)
    TTTFTTTTFF
    TTFFFFFFTF
    TFTTTTTTFF
    TFFTTTTTFF
    FTTFTTTTFF
    FTFFFTFTFF
    FFTTTTTTFF
    FFFTTTTTFF

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

QUESTION Miscellaneous Exercise 1 | Q 4.14 | Page 33

Using the truth table, prove the following logical equivalence.

  1. \( p \land (q \lor r) \equiv (p \land q) \lor (p \land r) \)
  2. \( [(\sim p \lor q) \lor (p \lor q)] \land r \equiv r \)
  3. \( p \land (\sim p \lor q) \equiv p \land q \)
  4. \( p \leftrightarrow q \equiv \sim (p \land \sim q) \land \sim (q \land \sim p) \)
  5. \( \sim p \land q \equiv (p \lor q) \land \sim p \)
SOLUTION
  1. \( p \land (q \lor r) \equiv (p \land q) \lor (p \land r) \) (Distributive Law)
    pqr\(q \lor r\)\(p \land (q \lor r)\) (LHS)\(p \land q\)\(p \land r\)\((p \land q) \lor (p \land r)\) (RHS)
    TTTTTTTT
    TTFTTTFT
    TFTTTFTT
    TFFFFFFF
    FTTTFFFF
    FTFTFFFF
    FFTTFFFF
    FFFFFFFF

    Columns for LHS and RHS are identical. Hence, the equivalence is proved.

  2. \( [(\sim p \lor q) \lor (p \lor q)] \land r \equiv r \)

    Let's simplify LHS first: \((\sim p \lor q) \lor (p \lor q) \equiv (\sim p \lor p) \lor (q \lor q) \equiv T \lor q \equiv T\). So, LHS becomes \(T \land r \equiv r\). Thus, LHS \(\equiv\) RHS.

    Using truth table for \([(\sim p \lor \sim q) \lor (p \lor q)] \land r \equiv r\) (as likely intended correction of `~(p ∨ q)` to `~p ∨ ~q` or considering it as is): Original form in prompt: `[~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r`. Let \(A = p \lor q\). Then the expression is \((\sim A \lor A) \land r\). Since \(\sim A \lor A\) is a tautology (T), this simplifies to \(T \land r \equiv r\). The equivalence holds. Columns for \(r\) and \([(\sim (p \lor q)) \lor (p \lor q)] \land r\) will be identical.

    pqr (RHS)\(p \lor q\)\(\sim(p \lor q)\)\(\sim(p \lor q) \lor (p \lor q)\)LHS
    TTTTFTT
    TTFTFTF
    TFTTFTT
    TFFTFTF
    FTTTFTT
    FTFTFTF
    FFTFTTT
    FFFFTTF

    Columns for LHS and RHS (column r) are identical. Hence, the equivalence is proved.

  3. \( p \land (\sim p \lor q) \equiv p \land q \) (Absorption Law variation)
    pq\(\sim p\)\(\sim p \lor q\)\(p \land (\sim p \lor q)\) (LHS)\(p \land q\) (RHS)
    TTFTTT
    TFFFFF
    FTTTFF
    FFTTFF

    Columns for LHS and RHS are identical. Hence, the equivalence is proved.

  4. \( p \leftrightarrow q \equiv \sim (p \land \sim q) \land \sim (q \land \sim p) \)
    pq\(p \leftrightarrow q\) (LHS)\(\sim q\)\(p \land \sim q\)\(\sim(p \land \sim q)\) (A)\(\sim p\)\(q \land \sim p\)\(\sim(q \land \sim p)\) (B)A \(\land\) B (RHS)
    TTTFFTFFTT
    TFFTTFFFTF
    FTFFFTTTFF
    FFTTFTTFTT

    Columns for LHS and RHS are identical. Hence, the equivalence is proved.

  5. \( \sim p \land q \equiv (p \lor q) \land \sim p \)
    pq\(\sim p\)\(\sim p \land q\) (LHS)\(p \lor q\)\((p \lor q) \land \sim p\) (RHS)
    TTFFTF
    TFFFTF
    FTTTTT
    FFTFFF

    Columns for LHS and RHS are identical. Hence, the equivalence is proved.

QUESTION Miscellaneous Exercise 1 | Q 4.15 | Page 33

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

  1. If \(2 + 5 = 10\), then \(4 + 10 = 20\).
  2. If a man is bachelor, then he is happy.
  3. If I do not work hard, then I do not prosper.
SOLUTION
  1. Let \(p: 2 + 5 = 10\), \(q: 4 + 10 = 20\). Statement: \(p \rightarrow q\).
    • Converse (\(q \rightarrow p\)): If \(4 + 10 = 20\), then \(2 + 5 = 10\).
    • Inverse (\(\sim p \rightarrow \sim q\)): If \(2 + 5 \neq 10\), then \(4 + 10 \neq 20\).
    • Contrapositive (\(\sim q \rightarrow \sim p\)): If \(4 + 10 \neq 20\), then \(2 + 5 \neq 10\).
  2. Let \(p\): A man is bachelor. \(q\): He is happy. Statement: \(p \rightarrow q\).
    • Converse (\(q \rightarrow p\)): If a man is happy, then he is bachelor.
    • Inverse (\(\sim p \rightarrow \sim q\)): If a man is not bachelor, then he is not happy.
    • Contrapositive (\(\sim q \rightarrow \sim p\)): If a man is not happy, then he is not bachelor.
  3. Let \(p\): I do not work hard. \(q\): I do not prosper. Statement: \(p \rightarrow q\).
    • Converse (\(q \rightarrow p\)): If I do not prosper, then I do not work hard.
    • Inverse (\(\sim p \rightarrow \sim q\)): If I work hard, then I prosper.
    • Contrapositive (\(\sim q \rightarrow \sim p\)): If I prosper, then I work hard.
QUESTION Miscellaneous Exercise 1 | Q 4.16 | Page 33

State the dual of the following statement by applying the principle of duality.

  1. \( (p \land \sim q) \lor (\sim p \land q) \equiv (p \lor q) \land \sim (p \land q) \)
  2. \( p \lor (q \lor r) \equiv \sim [(p \land q) \lor (r \lor s)] \)
  3. 2 is even number or 9 is a perfect square.
SOLUTION
  1. Dual: \( (p \lor \sim q) \land (\sim p \lor q) \equiv (p \land q) \lor \sim (p \lor q) \)
  2. Dual: \( p \land (q \land r) \equiv \sim [(p \lor q) \land (r \land s)] \)
  3. Dual: 2 is even number and 9 is a perfect square.
QUESTION Miscellaneous Exercise 1 | Q 4.17 | Page 33

Rewrite the following statement without using the connective ‘If ... then’.

  1. If a quadrilateral is rhombus then it is not a square.
  2. If \(10 - 3 = 7\) then \(10 \times 3 \neq 30\).
  3. If it rains then the principal declares a holiday.
SOLUTION

Using the equivalence \(p \rightarrow q \equiv \sim p \lor q\):

  1. A quadrilateral is not a rhombus or it is not a square.
  2. \(10 - 3 \neq 7\) or \(10 \times 3 \neq 30\).
  3. It does not rain or the principal declares a holiday.
QUESTION Miscellaneous Exercise 1 | Q 4.18 | Page 33

Write the dual of the following.

  1. \( (\sim p \land q) \lor (p \land \sim q) \lor (\sim p \land \sim q) \)
  2. \( (p \land q) \land r \equiv p \land (q \land r) \)
  3. \( p \lor (q \land r) \equiv (p \lor q) \land (p \lor r) \)
  4. \( \sim (p \lor q) \equiv \sim p \land \sim q \)
SOLUTION
  1. Dual: \( (\sim p \lor q) \land (p \lor \sim q) \land (\sim p \lor \sim q) \)
  2. Dual: \( (p \lor q) \lor r \equiv p \lor (q \lor r) \)
  3. Dual: \( p \land (q \lor r) \equiv (p \land q) \lor (p \land r) \)
  4. Dual: \( \sim (p \land q) \equiv \sim p \lor \sim q \)
QUESTION Miscellaneous Exercise 1 | Q 4.19 | Page 33

Consider the following statements.
i. If D is dog, then D is very good.
ii. If D is very good, then D is dog.
iii. If D is not very good, then D is not a dog.
iv. If D is not a dog, then D is not very good.
Identify the pairs of statements having the same meaning. Justify.

SOLUTION

Let \(p\): D is dog.
Let \(q\): D is very good.

The statements in symbolic form are:

  • i. \(p \rightarrow q\) (Original Statement)
  • ii. \(q \rightarrow p\) (Converse of i)
  • iii. \(\sim q \rightarrow \sim p\) (Contrapositive of i, Converse of iv, Inverse of ii)
  • iv. \(\sim p \rightarrow \sim q\) (Inverse of i, Contrapositive of ii, Converse of iii)

Pairs of statements having the same meaning (logically equivalent):

  • Pair 1: Statement (i) and Statement (iii)
    (\(p \rightarrow q \equiv \sim q \rightarrow \sim p\))
  • Pair 2: Statement (ii) and Statement (iv)
    (\(q \rightarrow p \equiv \sim p \rightarrow \sim q\))
QUESTION Miscellaneous Exercise 1 | Q 4.20 | Page 33

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

  1. All men are mortal.
  2. Some persons are not politician.
  3. Some members of the present Indian cricket team are not committed. (Corrected original: ...cricket are not committed)
  4. No child is an adult.
SOLUTION
  1. All men are mortal.
    Let U be the set of all human beings.
    Let M be the set of all men.
    Let R be the set of all mortals.
    The statement means M \(\subset\) R.
    Venn diagram for All men are mortal
  2. Some persons are not politician.
    Let U be the set of all human beings.
    Let P be the set of all persons.
    Let L be the set of all politicians.
    The statement means P \(\not\subseteq\) L, or P \(\cap\) L' \(\neq \emptyset\). (There exists at least one person who is not a politician).
    Venn diagram for Some persons are not politician
    (The shaded area X-Y \(\neq \emptyset\), assuming X is Persons and Y is Politicians).
  3. Some members of the present Indian cricket team are not committed.
    Let U be the set of all people.
    Let C be the set of members of the present Indian cricket team.
    Let M be the set of committed people.
    The statement means C \(\not\subseteq\) M, or C \(\cap\) M' \(\neq \emptyset\).
    Venn diagram for Some members of Indian cricket team are not committed
    (The shaded area M-C \(\neq \emptyset\) if M is 'members' and C is 'committed'. Or if C is 'members' and M is 'committed', then some part of C is outside M.)
  4. No child is an adult.
    Let U be the set of all human beings.
    Let C be the set of all children.
    Let A be the set of all adults.
    The statement means C \(\cap\) A = \(\emptyset\).
    Venn diagram for No child is an adult
QUESTION Miscellaneous Exercise 1 | Q 4.21 | Page 34

If \(A = \{2, 3, 4, 5, 6, 7, 8\}\), determine the truth value of the following statement.

  1. \( \exists x \in A \), such that \(3x + 2 > 9\)
  2. \( \forall x \in A, x^2 < 18 \)
  3. \( \exists x \in A \), such that \(x + 3 < 11 \)
  4. \( \forall x \in A, x^2 + 2 \ge 5 \)
SOLUTION

\(A = \{2, 3, 4, 5, 6, 7, 8\}\)

  1. \( \exists x \in A \), such that \(3x + 2 > 9\). Consider \(x=3 \in A\). \(3(3) + 2 = 9 + 2 = 11\). Since \(11 > 9\), the condition is satisfied for \(x=3\). Thus, the statement is True. Truth value: T.
  2. \( \forall x \in A, x^2 < 18 \). Consider \(x=5 \in A\). \(x^2 = 5^2 = 25\). Since \(25 \not< 18\), the condition is not satisfied for \(x=5\). Thus, the statement is False. Truth value: F.
  3. \( \exists x \in A \), such that \(x + 3 < 11 \). Consider \(x=2 \in A\). \(2 + 3 = 5\). Since \(5 < 11\), the condition is satisfied for \(x=2\). Thus, the statement is True. Truth value: T.
  4. \( \forall x \in A, x^2 + 2 \ge 5 \). Let's check for all \(x \in A\): For \(x=2\), \(2^2+2 = 4+2=6 \ge 5\) (True) For \(x=3\), \(3^2+2 = 9+2=11 \ge 5\) (True) For \(x=4\), \(4^2+2 = 16+2=18 \ge 5\) (True) For \(x=5\), \(5^2+2 = 25+2=27 \ge 5\) (True) For \(x=6\), \(6^2+2 = 36+2=38 \ge 5\) (True) For \(x=7\), \(7^2+2 = 49+2=51 \ge 5\) (True) For \(x=8\), \(8^2+2 = 64+2=66 \ge 5\) (True) Since the condition \(x^2+2 \ge 5\) is true for all \(x \in A\). Thus, the statement is True. Truth value: T.
QUESTION Miscellaneous Exercise 1 | Q 4.22 | Page 34

Write the negation of the following statement.

  1. 7 is prime number and Taj Mahal is in Agra.
  2. \(10 > 5\) and \(3 < 8\)
  3. I will have tea or coffee.
  4. \( \forall n \in N, n + 3 > 9 \)
  5. \( \exists x \in A \), such that \(x + 5 < 11 \) (Corrected original: ∃ n ∈ A)
SOLUTION
  1. Let \(p\): 7 is prime number. \(q\): Taj Mahal is in Agra. Statement: \(p \land q\). Negation: \(\sim(p \land q) \equiv \sim p \lor \sim q\). "7 is not a prime number or Taj Mahal is not in Agra."
  2. Let \(p: 10 > 5\). \(q: 3 < 8\). Statement: \(p \land q\). Negation: \(\sim(p \land q) \equiv \sim p \lor \sim q\). "\(10 \le 5\) or \(3 \ge 8\)."
  3. Let \(p\): I will have tea. \(q\): I will have coffee. Statement: \(p \lor q\). Negation: \(\sim(p \lor q) \equiv \sim p \land \sim q\). "I will not have tea and I will not have coffee." (Or: "I will have neither tea nor coffee.")
  4. Statement: \( \forall n \in N, n + 3 > 9 \). Negation: \( \exists n \in N \), such that \(\sim(n + 3 > 9)\), which is \( \exists n \in N \), such that \(n + 3 \le 9 \).
  5. Statement: \( \exists x \in A \), such that \(x + 5 < 11 \). Negation: \( \forall x \in A \), \(\sim(x + 5 < 11)\), which is \( \forall x \in A \), \(x + 5 \ge 11 \).
HSC Mathematics Full Solution