Maharashtra Board Class 12 Maths Part 1 Chapter 1 Mathematical Logic Miscellaneous Solutions

(I) Choose the Correct Alternative

Question 1. Which of the following is not a statement?
(a) Smoking is injurious to health
(b) 2 + 2 = 4
(c) Please, give me your pen
(d) Truth is beauty
Answer: (c) Please, give me your pen
In simple words: A statement must be a declarative sentence that is either definitely true or definitely false. A request like "Please, give me your pen" cannot be declared true or false, so it is not a statement.

🎯 Exam Tip: Remember that sentences expressing requests, commands, questions, or exclamations are never considered mathematical statements.

Question 1. Which of the following is not a statement?
(c) 2 is only even prime number
(d) Come here
Answer: (d) Come here
In simple words: A mathematical statement must be clearly true or false. A command like "Come here" cannot be declared true or false, so it is not a statement.

🎯 Exam Tip: Sentences that are orders, requests, questions, or exclamations are never considered statements in mathematical logic.

Question 2. Which of the following is an open statement?
(a) x is a natural number
(b) Give me a glass of water
(c) Wish you best of luck
(d) Good morning to all
Answer: (a) x is a natural number
In simple words: An open statement is a sentence whose truth depends on a variable, like "x", which can change depending on what number you plug in.

🎯 Exam Tip: Look for variables like x or y to easily identify open sentences in logic questions.

Question 3. Let \( p \wedge (q \vee r) = (p \wedge q) \vee (p \wedge r) \). Then this law is known as
(a) Commutative law
(b) Associative law
(c) De Morgan’s law
(d) Distributive law
Answer: (d) Distributive law
In simple words: This law shows how the "and" operator distributes over the "or" operator, just like multiplication distributes over addition in regular math.

🎯 Exam Tip: Memorize the standard logical laws like Distributive, Associative, and De Morgan's laws as they are frequently tested in exams.

Question 4. The false statement in the following is:
(a) \( p \wedge (\sim p) \) is a contradiction
(b) \( (p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p) \) is a contradiction
(c) \( \sim(\sim p) \leftrightarrow p \) is a tautology
(d) \( p \vee (\sim p) \leftrightarrow p \) is a tautology
Answer: (b) \( (p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p) \) is a contradiction
In simple words: A conditional statement and its contrapositive always have the exact same truth value, so their biconditional is a tautology (always true), not a contradiction.

🎯 Exam Tip: Remember that \( p \rightarrow q \) is logically equivalent to \( \sim q \rightarrow \sim p \), which makes their biconditional relation always true.

Question 5. Consider the following three statements
p : 2 is an even number.
q : 2 is a prime number.

Question 5. Let r : Sum of two prime numbers is always even. Then, the symbolic statement \( (p \wedge q) \rightarrow \sim r \) means:
(a) 2 is an even and prime number and the sum of two prime numbers is always even.
(b) 2 is an even and prime number and the sum of two prime numbers is not always even.
(c) If 2 is an even and prime number, then the sum of two prime numbers is not always even.
(d) If 2 is an even and prime number, then the sum of two prime numbers is also even.
Answer: (c) If 2 is an even and prime number, then the sum of two prime numbers is not always even.
In simple words: The symbol \( \rightarrow \) represents an "if-then" relationship, and \( \sim r \) means the opposite of statement r, so the entire expression translates to "if p and q, then not r".

🎯 Exam Tip: Identify the main logical connective first; here, the conditional arrow \( \rightarrow \) indicates that the sentence must be in the "If... then..." format.

Question 6. If p : He is intelligent, q : He is strong. Then, symbolic form of statement: ‘It is wrong that, he is intelligent or strong’ is
(a) \( \sim p \vee \sim p \)
(b) \( \sim(p \wedge q) \)
(c) \( \sim(p \vee q) \)
(d) \( p \vee \sim q \)
Answer: (c) \( \sim(p \vee q) \)
In simple words: "It is wrong that" means we negate the entire statement using a negation sign \( \sim \) outside the parentheses, and "or" is represented by the \( \vee \) symbol.

🎯 Exam Tip: Translate phrases like "It is wrong that" or "It is false that" as a negation symbol placed outside brackets containing the rest of the compound statement.

Question 7. The negation of the proposition ‘If 2 is prime, then 3 is odd’, is
(a) If 2 is not prime, then 3 is not odd
(b) 2 is prime and 3 is not odd
(c) 2 is not prime and 3 is odd
(d) If 2 is not prime, then 3 is odd
Answer: (b) 2 is prime and 3 is not odd
In simple words: The negation of an "if p, then q" statement is "p and not q", which means the first part is true but the second part is false.

🎯 Exam Tip: Remember the standard equivalence \( \sim(p \rightarrow q) \equiv p \wedge \sim q \). The negation of a conditional statement is never another conditional statement.

Question 8. The statement \( (\sim p \wedge q) \vee \sim q \) is
(a) a tautology
(b) a contradiction
(c) equivalent to \( \sim p \vee \sim q \)
(d) equivalent to \( \sim p \wedge \sim q \)
Answer: (c) equivalent to \( \sim p \vee \sim q \)
In simple words: By distributing the \( \vee \sim q \) inside, we get \( (\sim p \vee \sim q) \wedge (q \vee \sim q) \). Since \( q \vee \sim q \) is always true, the expression simplifies directly to \( \sim p \vee \sim q \).

🎯 Exam Tip: Use distributive laws to simplify complex logical expressions quickly instead of spending time drawing a full truth table.

Question 8. The expression \( (\sim p \wedge q) \vee \sim q \) is logically equivalent to
(a) \( p \vee q \)
(b) \( p \wedge q \)
(c) \( \sim(p \vee q) \)
(d) \( \sim(p \wedge q) \)
Answer: (d) \( \sim(p \wedge q) \)
Simplification steps:
\( (\sim p \wedge q) \vee \sim q \equiv (\sim p \vee \sim q) \wedge (q \vee \sim q) \)
\( \equiv (\sim p \vee \sim q) \wedge t \)
\( \equiv \sim p \vee \sim q \)
\( \equiv \sim(p \wedge q) \)
In simple words: We can distribute the OR operation over the AND operation, simplify the term with q, and then use De Morgan's Law to get the final simplified form.

🎯 Exam Tip: Remember that \( q \vee \sim q \) is always true (\( t \)), which helps eliminate terms quickly during simplification.

Question 9. Which of the following is always true?
(a) \( \sim(p \rightarrow q) \equiv \sim q \rightarrow \sim p \)
(b) \( \sim(p \vee q) \equiv \sim p \vee \sim q \)
(c) \( \sim(p \rightarrow q) \equiv p \wedge \sim q \)
(d) \( \sim(p \wedge q) \equiv \sim p \wedge \sim q \)
Answer: (c) \( \sim(p \rightarrow q) \equiv p \wedge \sim q \)
In simple words: The negation of a conditional statement "if p then q" is "p and not q". For example, the negation of "If it rains, I will stay home" is "It rains and I do not stay home."

🎯 Exam Tip: Memorize the standard negation rules for conditional statements, as they are frequently asked in exams.

Question 10. \( \sim(p \vee q) \vee (\sim p \wedge q) \) is logically equivalent to
(a) \( \sim p \)
(b) \( p \)
(c) \( q \)
(d) \( \sim q \)
Answer: (a) \( \sim p \)
Simplification steps:
\( \sim(p \vee q) \vee (\sim p \wedge q) \equiv (\sim p \wedge \sim q) \vee (\sim p \wedge q) \)
\( \equiv \sim p \wedge (\sim q \vee q) \)
\( \equiv \sim p \wedge t \)
\( \equiv \sim p \)
In simple words: By breaking down the first part using De Morgan's law, we can factor out "not p" from both parts, which simplifies the remaining terms to always true, leaving just "not p".

🎯 Exam Tip: Look for common terms like \( \sim p \) that can be factored out using the distributive law to simplify complex logical expressions.

Question 11. If p and q are two statements, then \( (p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p) \) is
(a) contradiction
(b) tautology
(c) contingency
(d) none of the options
Answer: (b) tautology
In simple words: A conditional statement is always logically equivalent to its contrapositive, so their biconditional relationship is always true (a tautology).

🎯 Exam Tip: Any biconditional statement \( A \leftrightarrow B \) where \( A \equiv B \) is always a tautology because both sides always share the same truth value.

Question 12. If p is the sentence ‘This statement is false’, then
(a) truth value of p is T
(b) truth value of p is F
(c) p is both true and false
(d) p is neither true nor false
Answer: (d) p is neither true nor false
In simple words: This is a famous self-contradicting statement. If we assume it is true, then it must be false; if we assume it is false, then it must be true. Because it creates a loop of contradiction, it cannot be labeled as simply true or false.

🎯 Exam Tip: Statements that refer to their own truth value and contradict themselves are logical paradoxes and do not have a standard truth value of T or F.

Question 13. Conditional \( p \rightarrow q \) is equivalent to
(a) \( p \rightarrow \sim q \)
(b) \( \sim p \vee q \)
(c) \( \sim p \rightarrow \sim q \)
(d) \( p \vee \sim q \)
Answer: (b) \( \sim p \vee q \)
In simple words: Saying "If it rains, then the ground is wet" means the same as saying "Either it does not rain, or the ground is wet."

🎯 Exam Tip: Memorize the conditional law \( p \rightarrow q \equiv \sim p \vee q \) as it is one of the most frequently tested equivalences in mathematical logic.

Question 14. Negation of the statement ‘This is false or That is true’ is
(a) That is true or This is false
(b) That is true and This is false
(c) This is true and That is false
(d) That is false and That is true
Answer: (c) This is true and That is false
In simple words: To find the opposite of an "or" statement, you must change "or" to "and" and write the opposite of both parts. So, "This is false" becomes "This is true", and "That is true" becomes "That is false".

🎯 Exam Tip: Always apply De Morgan's Law \( \sim(A \vee B) \equiv \sim A \wedge \sim B \) by negating both statements and changing the disjunction (or) to a conjunction (and).

(Ii) Fill in the Blanks:

Question 1. The statement \( q \rightarrow p \) is called as the ___________ of the statement \( p \rightarrow q \).
Answer: Converse. Swapping the hypothesis and conclusion of a conditional statement always yields its converse.
In simple words: When you swap the order of an "if-then" statement, the new statement is called its converse.

🎯 Exam Tip: Remember the order: \( p \rightarrow q \) is the conditional, \( q \rightarrow p \) is the converse, \( \sim p \rightarrow \sim q \) is the inverse, and \( \sim q \rightarrow \sim p \) is the contrapositive.

Question 2. Conjunction of two statements \( p \) and \( q \) is symbolically written as
Answer: \( p \wedge q \). This symbol represents the logical "AND" operator which connects two individual statements.
In simple words: To show that both statements must be true at the same time, we use the "AND" symbol, which looks like an upside-down 'v'.

🎯 Exam Tip: Do not confuse the conjunction symbol \( \wedge \) (AND) with the disjunction symbol \( \vee \) (OR).

Question 3. If \( p \vee q \) is true, then truth value of \( \sim p \vee \sim q \) is ___________
Answer: False. This specific outcome is determined under the assumption that both individual statements are true, which makes their negations false.
In simple words: If both parts are true, then making them both opposite (false) and joining them with "OR" will result in a false statement.

🎯 Exam Tip: When solving truth value problems, construct a quick truth table or test with specific truth values (like T and T) to verify the result.

Question 4. Negation of ‘some men are animal’ is ___________
Answer: All men are not animal. (OR No men are animals.) This logical negation completely denies the existence of any man being an animal.
In simple words: The opposite of saying "at least some are" is saying "absolutely none are".

🎯 Exam Tip: The negation of a quantified statement containing "some" always changes to "all... not" or "no".

Question 5. Truth value of if \( x = 2 \), then \( x^2 = -4 \) is ___________
Answer: False. Since the first part is true but the second part is false, the overall conditional statement becomes false.
In simple words: An "if-then" statement is false if the starting assumption is true but the final result turns out to be false.

🎯 Exam Tip: A conditional statement \( p \rightarrow q \) is only false when the antecedent \( p \) is true and the consequent \( q \) is false.

Question 6. Inverse of statement pattern \( p \rightarrow q \) is given by ___________
Answer: \( \sim p \rightarrow \sim q \). This is formed by negating both the antecedent and the consequent of the original conditional statement.
In simple words: To find the inverse, you simply add "not" to both parts of the original "if-then" statement.

🎯 Exam Tip: Keep the order of the statements the same for the inverse, and only negate them; do not swap them.

Question 7. \( p \leftrightarrow q \) is false when p and q have ___________ truth values.
Answer: Different. This is because a biconditional statement is only true when both component statements share the same truth value.
In simple words: A double-arrow statement is false if one part is true and the other part is false.

🎯 Exam Tip: Remember that biconditional (if and only if) statements require both sides to match in truth value to be true.

Question 8. Let \( p \) : The problem is easy. \( r \) : It is not challenging. Then verbal form of \( \sim p \rightarrow r \) is ___________
Answer: If the problem is not easy, then it is not challenging. This conditional statement directly translates the negation of p and the statement r.
In simple words: We replace the symbols with their word meanings, where the tilde means "not" and the arrow means "if... then...".

🎯 Exam Tip: Always translate the negation symbol first before writing down the conditional "if-then" structure.

Question 9. Truth value of \( 2 + 3 = 5 \) if and only if \( -3 > -9 \) is ___________
Answer: T [Hint: \( T \leftrightarrow T = T \)]. Since both individual statements are true, their biconditional combination is also true.
In simple words: Since 2 + 3 = 5 is true, and -3 is indeed greater than -9, both parts are true, making the whole statement true.

🎯 Exam Tip: Break down biconditional statements into two separate parts and find the truth value of each part first.

(III) State whether each of the following is True or False:

Question 1. Truth value of \( 2 + 3 < 6 \) is F.
Answer: False. The statement is false because 2 + 3 equals 5, which is indeed less than 6, making its actual truth value T.
In simple words: The question claims the truth value is False, but since 5 is less than 6, the statement is actually True. Therefore, the claim itself is False.

🎯 Exam Tip: Read carefully: if a question asks if a statement's truth value is 'F', you must evaluate the statement first and then check if it matches 'F'.

Question 2. There are 24 months in a year is a statement.
Answer: True. It is a statement because it is a declarative sentence that has a definite truth value, even though that truth value happens to be false.
In simple words: A statement in logic is any sentence that is either completely true or completely false. Since we can definitely say this sentence is false, it counts as a statement.

🎯 Exam Tip: Remember that a sentence does not have to be factually correct to be a mathematical statement; it just needs to have a clear true or false value.

Question 3. \( p \wedge q \) has truth value F if both p and q have truth value F.
Answer: False. A conjunction statement is false if any of its parts are false, meaning it does not require both to be false to result in a false value.
In simple words: An "and" statement is false if even one part is false, so it is false in multiple cases, not just when both parts are false.

🎯 Exam Tip: For conjunction (AND) statements, remember that it only takes one false component to make the entire statement false.

Question 4. The negation of \( 10 + 20 = 30 \) is, it is false that \( 10 + 20 \neq 30 \).
Answer: False. The negation of \( 10 + 20 = 30 \) is simply \( 10 + 20 \neq 30 \), or "it is false that \( 10 + 20 = 30 \)". Adding "it is false that" to an already negated statement creates a double negation, which changes the meaning back to the original statement.
In simple words: Negating a statement means saying it is not true. If we say "it is false that they are not equal," we are actually saying they are equal, which is not the correct negation.

🎯 Exam Tip: Remember that a double negative cancels out. To find the negation of \( A = B \), simply write \( A \neq B \) without adding extra "it is false that" phrases.

Question 5. Dual of \( (p \wedge \sim q) \vee t \) is \( (p \vee \sim q) \vee c \).
Answer: False. To find the dual of a statement, we replace \( \wedge \) with \( \vee \), \( \vee \) with \( \wedge \), \( t \) with \( c \), and \( c \) with \( t \). Therefore, the correct dual of \( (p \wedge \sim q) \vee t \) is \( (p \vee \sim q) \wedge c \).
In simple words: When finding the dual, every "and" becomes "or", every "or" becomes "and", and "true" becomes "contradiction". Here, the middle "or" symbol was not changed to "and", making the statement false.

🎯 Exam Tip: Carefully check every single connective and constant when writing a dual; it is extremely common to miss one in the middle of a bracket.

Question 6. Dual of ‘John and Ayub went to the forest’ is ‘John or Ayub went to the forest.’
Answer: True. In mathematical logic, the word 'and' represents the conjunction operator \( \wedge \), and 'or' represents the disjunction operator \( \vee \). Replacing 'and' with 'or' correctly gives the dual of the statement.
In simple words: To find the dual of a word statement, we simply swap the word "and" with "or". Since this statement does exactly that, it is true.

🎯 Exam Tip: For verbal statements, identify the logical connectives like 'and' and 'or' first, then swap them to find the dual statement easily.

Question 7. ‘His birthday is on 29th February’ is not a statement.
Answer: True. The sentence contains the pronoun 'His', which makes its truth value depend on who 'he' is. Since its truth value varies depending on the person referred to, it is an open sentence and therefore not a logical statement.
In simple words: A sentence is only a statement if we can definitely say it is true or false. Because we do not know who "he" is, we cannot decide if this is true or false, so it is not a statement.

🎯 Exam Tip: Any sentence containing pronouns like 'he', 'she', 'it', or variables like 'x' and 'y' without specified values is an open sentence, not a statement.

Question 8. \( x^2 = 25 \) is true statement.
Answer: False. The equation \( x^2 = 25 \) is an open sentence because its truth value depends on the value assigned to the variable \( x \). It is true for \( x = 5 \) or \( x = -5 \), but false for other values, meaning it cannot be classified as a statement.
In simple words: An equation with a variable like \( x \) is not a statement because we cannot say it is true or false until we know what \( x \) is.

🎯 Exam Tip: Mathematical equations with variables are open sentences and are not considered statements unless the value of the variable is predefined.

Question 9. The truth value of ‘\( \sqrt{5} \) is not an irrational number’ is T.
Answer: False. Since \( \sqrt{5} \) is indeed an irrational number, the statement claiming it is "not an irrational number" is false, meaning its truth value is F, not T.
In simple words: We know that \( \sqrt{5} \) is an irrational number because it cannot be written as a simple fraction. Therefore, saying it is "not" irrational is incorrect, so the truth value is False.

🎯 Exam Tip: Always determine the actual mathematical fact first, then compare it to the given statement to find the correct truth value.

Question 10. \( p \wedge t \equiv p \).
Answer: True. According to the Identity Law in mathematical logic, the conjunction of any statement \( p \) with a tautology \( t \) always has the same truth value as \( p \) itself.
In simple words: Combining any statement with a true statement using "and" doesn't change its value. If \( p \) is true, the result is true; if \( p \) is false, the result is false.

🎯 Exam Tip: Memorize the Identity Laws: \( p \wedge t \equiv p \) and \( p \vee c \equiv p \) as they are frequently asked in true/false and simplification questions.

(IV) Solve the Following:

Question 1. State which of the following sentences are statements in logic:
(i) Ice cream Sundaes are my favourite.
(ii) \( x + 3 = 8 \), \( x \) is variable.
(iii) Read a lot to improve your writing skill.
(iv) \( z \) is a positive number.
(v) \( (a + b)^2 = a^2 + 2ab + b^2 \) for all \( a, b \in R \).
(vi) \( (2 + 1)^2 = 9 \).
(vii) Why are you sad?
(viii) How beautiful the flower is!
(ix) The square of any odd number is even.
Answer:
(i) It is a statement.
(ii) It is a statement.
(iii) It is an imperative sentence, hence it is not a statement.
(iv) It is an open sentence, hence it is not a statement.
(v) It is a statement.
(vi) It is a statement.
(vii) It is an interrogative sentence, hence it is not a statement.
(viii) It is an exclamatory sentence, hence it is not a statement.
(ix) It is a statement.
In simple words: A statement in logic is any sentence that is clearly either true or false. Sentences that ask questions, give orders, express strong feelings, or contain unknown variables whose values aren't specified are not statements.

🎯 Exam Tip: Remember that sentences expressing requests, commands, questions, or exclamations are never statements in mathematical logic. Always state the reason clearly if a sentence is not a statement.

Question 1. State whether the following sentences are statements:
(x) All integers are natural numbers.
(xi) If \( x \) is a real number, then \( x^2 \ge 0 \).
(xii) Do not come inside the room.
(xiii) What a horrible sight it was!
Answer:
(x) It is a statement.
(xi) It is a statement.
(xii) It is an imperative sentence, hence it is not a statement.
(xiii) It is an exclamatory sentence, hence it is not a statement.
In simple words: A mathematical statement must be a declarative sentence that is either completely true or completely false. Commands, requests, and exclamations do not count as statements.

🎯 Exam Tip: Identify the type of sentence first; sentences that express orders, requests, or strong emotions are never mathematical statements.

Question 2. Which of the following sentences are statements? In case of a statement, write down the truth value:
(i) What is a happy ending?
(ii) The square of every real number is positive.
(iii) Every parallelogram is a rhombus.
(iv) \( a^2 - b^2 = (a + b)(a - b) \) for all \( a, b \in R \).
Answer:
(i) It is an interrogative sentence, hence it is not a statement.
(ii) It is a statement that is false, hence its truth value is F.
(iii) It is a statement that is true, hence its truth value is T.
(iv) It is a mathematical identity that is true, hence its truth value is T.
In simple words: Questions are not statements. For actual statements, we write 'T' if the statement is mathematically correct and 'F' if it is incorrect.

🎯 Exam Tip: For algebraic identities that hold true for all real numbers, the truth value is always T. Be careful with definitions of geometric shapes when determining truth values.

Question 2. State whether the following sentences are statements. Justify. In case of a statement, state its truth value.
(v) Please carry out my instruction.
(vi) The Himalayas is the highest mountain range.
(vii) \( (x - 2)(x - 3) = x^2 - 5x + 6 \) for all \( x \in R \).
(viii) What are the causes of rural unemployment?
(ix) \( 0! = 1 \).
(x) The quadratic equation \( ax^2 + bx + c = 0 \) (\( a \neq 0 \)) always has two real roots.
Answer:
(v) It is an imperative sentence, hence it is not a statement.
(vi) It is a statement that is true, hence its truth value is T.
(vii) It is a mathematical identity that is true, hence its truth value is T.
(viii) It is an interrogative sentence, hence it is not a statement.
(ix) It is a statement that is true, hence its truth value is T.
(x) It is a statement that is false, hence its truth value is F.
In simple words: Statements are sentences that are either completely true or completely false. Commands, questions, and false mathematical generalizations are handled accordingly based on whether they declare a clear, unchanging truth.

🎯 Exam Tip: Remember that imperative, interrogative, and exclamatory sentences are never mathematical statements. Always state the reason clearly before writing the truth value.

Question 3. Assuming the first statement as p and second as q, write the following statements in symbolic form:
(i) The Sun has set and Moon has risen.
(ii) Mona likes Mathematics and Physics.
Answer:
(i) Let \( p \) : The Sun has set.
\( q \) : Moon has risen.
Then the symbolic form of the given statement is \( p \wedge q \).

(ii) Let \( p \) : Mona likes Mathematics.
\( q \) : Mona likes Physics.
Then the symbolic form of the given statement is \( p \wedge q \).
In simple words: To write statements in symbols, we represent each simple sentence with a letter like p or q, and use the symbol \( \wedge \) to represent the word 'and'.

🎯 Exam Tip: Clearly define the component statements \( p \) and \( q \) before writing the final symbolic form to ensure you get full marks.

Question 1. Express the following statements in symbolic form:
(iii) 3 is a prime number if 3 is a perfect square number.
(iv) Kavita is brilliant and brave.
(v) If Kiran drives a car, then Sameer will walk.
(vi) The necessary condition for the existence of a tangent to the curve of the function is continuity.
(vii) To be brave is necessary and sufficient condition to climb Mount Everest.
(viii) \( x^3 + y^3 = (x + y)^3 \), iff \( xy = 0 \).
Answer:
(iii) Let p : 3 be a prime number.
q : 3 is a perfect square number.
Then the symbolic form of the given statement is \( p \leftrightarrow q \).

(iv) Let p : Kavita is brilliant.
q : Kavita is brave.
Then the symbolic form of the given statement is \( p \wedge q \).

(v) Let p : Kiran drives a car.
q : Sameer will walk.
Then the symbolic form of the given statement is \( p \rightarrow q \).

(vi) The given statement can be written as: 'If the function is continuous, then the tangent to the curve exists.'
Let p : The function is continuous.
q : Tangent to the curve exists.
Then the symbolic form of the given statement is \( p \rightarrow q \).

(vii) Let p : To be brave.
q : Climb Mount Everest.
Then the symbolic form of the given statement is \( p \leftrightarrow q \).

(viii) Let p : \( x^3 + y^3 = (x + y)^3 \).
q : \( xy = 0 \).
Then the symbolic form of the given statement is \( p \leftrightarrow q \). These symbolic representations help simplify complex logical arguments in mathematics.
In simple words: We can turn everyday sentences into math symbols by using letters like p and q for simple statements, and symbols like arrows or 'and' signs to connect them. This makes it easier to see the logical structure of what we are saying.

🎯 Exam Tip: Identify the connective words like 'if... then', 'and', or 'iff' first, as they directly tell you which logical symbol (\( \rightarrow \), \( \wedge \), or \( \leftrightarrow \)) to use.

Question (ix) Express the following statement in symbolic form: "The drug is effective though it has side effects."
Answer:
Let \( p \): The drug is effective.
\( q \): It has side effects.
Then the symbolic form of the given statement is \( p \wedge q \). This representation captures the logical connection between the two individual statements.
In simple words: The word 'though' acts like 'and' in logic, so we connect the two statements with the 'and' symbol (\( \wedge \)).

🎯 Exam Tip: Remember that words like 'though', 'but', 'still', and 'even though' translate to the logical conjunction 'and' (\( \wedge \)).

Question (x) Express the following statement in symbolic form: "If a real number is not rational, then it must be irrational."
Answer:
Let \( p \): A real number is not rational.
\( q \): It must be irrational.
Then the symbolic form of the given statement is \( p \rightarrow q \). This conditional statement shows the logical flow from the hypothesis to the conclusion.
In simple words: An 'if... then...' statement is a conditional statement, which we represent using the arrow symbol (\( \rightarrow \)).

🎯 Exam Tip: Identify the 'if' part as the antecedent \( p \) and the 'then' part as the consequent \( q \), and connect them with \( \rightarrow \).

Question (xi) Express the following statement in symbolic form: "It is not true that Ram is tall and handsome."
Answer:
Let \( p \): Ram is tall.
\( q \): Ram is handsome.
Then the symbolic form of the given statement is \( \sim(p \wedge q) \). This negation applies to the entire conjunction of the two traits.
In simple words: 'It is not true' means we put a negation sign (\( \sim \)) outside the brackets that hold the 'and' statement.

🎯 Exam Tip: When a negation applies to an entire combined statement, use parentheses to show that the negation symbol \( \sim \) affects the whole group.

Question (xii) Express the following statement in symbolic form: "Even though it is not cloudy, it is still raining."
Answer:
The given statement is equivalent to: "It is not cloudy and it is still raining."
Let \( p \): It is not cloudy.
\( q \): It is still raining.
Then the symbolic form of the given statement is \( p \wedge q \). This clearly shows both conditions occurring simultaneously.
In simple words: 'Even though' and 'still' work just like 'and', so we connect the two parts with the \( \wedge \) symbol.

🎯 Exam Tip: Always simplify the English sentence into standard logical connectives like 'and' or 'or' before writing the final symbolic form.

Question (xiii) Express the following statement in symbolic form: "It is not true that intelligent persons are neither polite nor helpful."
Answer:
Let \( p \): Intelligent persons are neither polite nor helpful.
Then the symbolic form of the given statement is \( \sim p \). This represents the negation of the entire statement about intelligent persons.
In simple words: Since the main statement is defined as \( p \), saying 'it is not true' simply negates it, giving us \( \sim p \).

🎯 Exam Tip: If the entire compound statement is already defined as a single variable \( p \), its negation is simply \( \sim p \).

Question (xiv) Express the following statement in symbolic form: "If the question paper is not easy, then we shall not pass."
Answer:
Let \( p \): The question paper is not easy.
Let \( q \): We shall not pass.
Then the symbolic form of the given statement is \( p \rightarrow q \). This conditional statement links the difficulty of the paper to the passing outcome.
In simple words: Since both parts contain 'not', we can define them directly as \( p \) and \( q \) and connect them with an arrow (\( \rightarrow \)) for 'if... then...'.

🎯 Exam Tip: You can define your simple statements with or without the negation; just make sure your symbolic form matches your definitions consistently.

Question 4. If \( p \) : Proof is lengthy, \( q \) : It is interesting. Express the following statements in symbolic form:
(i) Proof is lengthy and it is not interesting.
(ii) If the proof is lengthy, then it is interesting.
(iii) It is not true that the proof is lengthy but it is interesting.
(iv) It is interesting iff the proof is lengthy.
Answer:
(i) \( p \wedge \sim q \)
(ii) \( p \rightarrow q \)
(iii) \( \sim(p \wedge q) \)
(iv) \( q \leftrightarrow p \)
These symbolic representations help simplify complex logical statements into mathematical forms.
In simple words: We use symbols like \( \wedge \) for 'and', \( \rightarrow \) for 'if...then', \( \sim \) for 'not', and \( \leftrightarrow \) for 'if and only if' to write long sentences in a short mathematical way.

🎯 Exam Tip: Be careful with 'but' in logical statements; it is translated as 'and' (\( \wedge \)) in symbolic logic. Always identify the main connective first.

Question 5. Let \( p \) : Sachin win the match, \( q \) : Sachin is a member of the Rajya Sabha, \( r \) : Sachin is happy. Write the verbal statement for each of the following:
(i) \( (p \wedge q) \vee r \)
(ii) \( p \rightarrow r \)
(iii) \( \sim p \vee q \)
Answer:
(i) Sachin wins the match and he is a member of the Rajya Sabha or Sachin is happy.
(ii) If Sachin wins the match, then he is happy.
(iii) Sachin does not win the match or he is a member of the Rajya Sabha.
Translating these symbols back into words helps us understand the real-world meaning of logical expressions.
In simple words: Here, we are doing the opposite of the previous question by turning math symbols back into normal English sentences. We replace \( \wedge \) with 'and', \( \vee \) with 'or', and \( \rightarrow \) with 'if...then'.

🎯 Exam Tip: When writing verbal statements, ensure the sentence flows naturally in English while strictly maintaining the logical structure of the symbols.

Question 5. Express the following symbolic statements in words, where p: Sachin wins the match, q: Sachin is a member of the Rajya Sabha, and r: Sachin is happy:
(iv) \( p \rightarrow (q \vee r) \)
(v) \( p \rightarrow q \)
(vi) \( (p \wedge q) \wedge \sim r \)
(vii) \( \sim(p \vee q) \wedge r \)
Answer:
(iv) If Sachin wins the match, then he is a member of the Rajya Sabha or he is happy.
(v) If Sachin wins the match, then he is a member of the Rajya Sabha.
(vi) Sachin wins the match and he is a member of the Rajya Sabha but he is not happy. This demonstrates how multiple logical conditions can be combined using conjunctions and negations.
(vii) It is false that Sachin wins the match or he is a member of the Rajya Sabha but he is happy.
In simple words: We translate mathematical logic symbols back into English sentences. The symbol \( \rightarrow \) means "if... then", \( \wedge \) means "and" (or "but"), \( \vee \) means "or", and \( \sim \) means "not" or "it is false that".

🎯 Exam Tip: Remember that "but" is logically equivalent to "and" (\( \wedge \)) in mathematical statements.

Question 6. Determine the truth values of the following statements:
(i) \( 4 + 5 = 7 \) or \( 9 - 2 = 5 \).
(ii) If \( 9 > 1 \), then \( x^2 - 2x + 1 = 0 \) for \( x = 1 \).
(iii) \( x + y = 0 \) is the equation of a straight line if and only if \( y^2 = 4x \) is the equation of the parabola.
Answer:
(i)
Let \( p : 4 + 5 = 7 \)
\( q : 9 - 2 = 5 \)
Then the symbolic form of the given statement is \( p \vee q \).
The truth values of both \( p \) and \( q \) are F.
\( \therefore \) the truth value of \( p \vee q \) is F. [\( \text{F} \vee \text{F} \equiv \text{F} \)]

(ii)
Let \( p : 9 > 1 \)
\( q : x^2 - 2x + 1 = 0 \) for \( x = 1 \).
Then the symbolic form of the given statement is \( p \rightarrow q \).
The truth values of both \( p \) and \( q \) are T.
\( \dots \) the truth value of \( p \rightarrow q \) is T. [\( \text{T} \rightarrow \text{T} \equiv \text{T} \)]

(iii)
Let \( p : x + y = 0 \) is the equation of a straight line.
\( q : y^2 = 4x \) is the equation of the parabola.
Then the symbolic form of the given statement is \( p \leftrightarrow q \).
The truth values of both \( p \) and \( q \) are T because \( x + y = 0 \) represents a linear equation and \( y^2 = 4x \) represents a standard right-opening parabola. This shows how algebraic equations can be analyzed using formal mathematical logic.
\( \therefore \) the truth value of \( p \leftrightarrow q \) is T. [\( \text{T} \leftrightarrow \text{T} \equiv \text{T} \)]
In simple words: To find the truth value of a compound statement, first break it down into simple statements, find their individual truth values (True or False), and then use the truth tables of the logical connectives (like 'or', 'if... then', 'if and only if') to get the final result.

🎯 Exam Tip: Always write down the individual statements \( p \) and \( q \) clearly with their truth values before applying the logical connective rules to avoid simple calculation mistakes.

Question (iv). It is not true that 2 + 3 = 6 or 12 + 3 = 5.
Answer:
Let \( p \) : 2 + 3 = 6.
\( q \) : 12 + 3 = 5.
Then the symbolic form of the given statement is \( \sim(p \vee q) \).
The truth values of both \( p \) and \( q \) are F.

\( \implies \) the truth value of \( \sim(p \vee q) \) is T. [\( \sim(F \vee F) \equiv \sim F \equiv T \)]
In simple words: The statement says it is not true that either of the two false equations is correct. Since both equations are indeed false, the overall statement is true.

🎯 Exam Tip: Always write down the individual simple statements p and q with their truth values before finding the truth value of the compound statement.

Question 7. Assuming the following statements:
p : Stock prices are high.
q : Stocks are rising.
to be true, find the truth values of the following:
(i) Stock prices are not high or stocks are rising.
(ii) Stock prices are high and stocks are rising if and only if stock prices are high.
(iii) If stock prices are high, then stocks are not rising.
Answer:
(i) Since \( p \) and \( q \) are true (T):

\( \implies \) \( \sim p \) and \( \sim q \) are false (F).
The given statement in symbolic form is \( \sim p \vee q \).
Since \( \sim T \vee T \equiv F \vee T \equiv T \), the given statement is true. Hence, its truth value is 'T'.

(ii) The given statement in symbolic form is \( (p \wedge q) \leftrightarrow p \).
Since \( (T \wedge T) \leftrightarrow T \equiv T \leftrightarrow T \equiv T \), the given statement is true. Hence, its truth value is 'T'.

(iii) The given statement in symbolic form is \( p \rightarrow \sim q \).
Since \( T \rightarrow \sim T \equiv T \rightarrow F \equiv F \), the given statement is false. Hence, its truth value is 'F'.
In simple words: We assume both "stock prices are high" and "stocks are rising" are true. We then translate each compound sentence into logic symbols, substitute T or F, and calculate the final truth value.

🎯 Exam Tip: Clearly define the symbolic form first, then substitute the given truth values step-by-step to avoid errors in conditional and biconditional statements.

Question 7. State the symbolic form and determine the truth value of the following:
(iv) It is false that stocks are rising and stock prices are high.
(v) Stock prices are high or stocks are not rising iff stocks are rising.
Answer:
(iv) The given statement in symbolic form is \( \sim(q \wedge p) \).
Since, \( \sim(T \wedge T) \equiv \sim T \equiv F \), the given statement is false. Hence, its truth value is 'F'.

(v) The given statement in symbolic form is \( (p \vee \sim q) \leftrightarrow q \).
Since \( (T \vee \sim T) \leftrightarrow T \equiv (T \vee F) \leftrightarrow T \equiv T \leftrightarrow T \equiv T \), the given statement is true. Hence, its truth value is 'T'.
In simple words: We convert the sentences into math symbols using 'and', 'or', and 'if and only if'. By putting in the true (T) or false (F) values, we can calculate whether the whole statement is true or false.

🎯 Exam Tip: Always write the symbolic form first before substituting the truth values to avoid calculation errors.

Question 8. Rewrite the following statements without using conditional: [Hint: \( p \rightarrow q \equiv \sim p \vee q \)]
(i) If price increases, then demand falls.
(ii) If demand falls, then the price does not increase.
Answer:
Since, \( p \rightarrow q \equiv \sim p \vee q \), the given statements can be written as:
(i) Price does not increase or demand falls.
(ii) Demand does not fall or price does not increase.
In simple words: An "if-then" statement can be rewritten by negating the first part and combining it with the second part using "or".

🎯 Exam Tip: Remember the conditional equivalence formula \( p \rightarrow q \equiv \sim p \vee q \) as it is frequently asked in exams to rewrite conditional statements.

Question 9. If \( p, q, r \) are statements with truth values \( T, T, F \) respectively, determine the truth values of the following:
(i) \( (p \wedge q) \rightarrow \sim p \)
Answer:
Truth values of \( p, q, r \) are \( T, T, F \) respectively.
\( (p \wedge q) \rightarrow \sim p \equiv (T \wedge T) \rightarrow \sim T \)
\( \equiv T \rightarrow F \)
\( \equiv F \)
Hence, the truth value is 'F'.
In simple words: We substitute the given truth values into the expression and simplify step-by-step to find the final truth value.

🎯 Exam Tip: Substitute the truth values carefully step-by-step, working from the innermost parentheses outward to ensure accuracy.

Question 1. Find the truth values of the following statement patterns:
(ii) \( p \leftrightarrow (q \rightarrow \sim p) \)
(iii) \( (p \wedge \sim q) \vee (\sim p \wedge q) \)
(iv) \( \sim(p \wedge q) \rightarrow \sim(q \wedge p) \)
(v) \( \sim[(p \rightarrow q) \leftrightarrow (p \wedge \sim q)] \)
Answer:
(ii) \( p \leftrightarrow (q \rightarrow \sim p) \)
Solution:
\( p \leftrightarrow (q \rightarrow \sim p) \equiv T \leftrightarrow (T \rightarrow \sim T) \)
\( \equiv T \leftrightarrow (T \rightarrow F) \)
\( \equiv T \leftrightarrow F \)
\( \equiv F \)
Hence, the truth value of the given statement is false, i.e. F.

(iii) \( (p \wedge \sim q) \vee (\sim p \wedge q) \)
Solution:
\( (p \wedge \sim q) \vee (\sim p \wedge q) \equiv (T \wedge \sim T) \vee (\sim T \wedge T) \)
\( \equiv (T \wedge F) \vee (F \wedge T) \)
\( \equiv F \vee F \)
\( \equiv F \)
Hence, the truth value of the given statement is false, i.e. F.

(iv) \( \sim(p \wedge q) \rightarrow \sim(q \wedge p) \)
Solution:
\( \sim(p \wedge q) \rightarrow \sim(q \wedge p) \equiv \sim(T \wedge T) \rightarrow \sim(T \wedge T) \)
\( \equiv \sim T \rightarrow \sim T \)
\( \equiv F \rightarrow F \)
\( \equiv T \)
Hence, the truth value of the given statement is true, i.e. T.

(v) \( \sim[(p \rightarrow q) \leftrightarrow (p \wedge \sim q)] \)
Solution:
\( \sim[(p \rightarrow q) \leftrightarrow (p \wedge \sim q)] \)
\( \equiv \sim[(T \rightarrow T) \leftrightarrow (T \wedge \sim T)] \)
\( \equiv \sim[T \leftrightarrow (T \wedge F)] \)
\( \equiv \sim[T \leftrightarrow F] \)
\( \equiv \sim F \)
\( \equiv T \)
Hence, the truth value of the given statement is true, i.e. T.
In simple words: To find the truth value of a compound statement, we replace each letter with its true (T) or false (F) value. Then, we solve the brackets step-by-step just like normal math, using logical rules to get the final single T or F answer.

🎯 Exam Tip: Show every intermediate step of substitution and simplification to secure full marks. Be especially careful with the negation symbol outside brackets, applying it only after solving the entire bracket.

Question 10. Write the negations of the following:
(i) If \(\Delta\text{ABC}\) is not equilateral, then it is not equiangular.
(ii) Ramesh is intelligent and he is hard working.
(iii) A angle is a right angle if and only if it is of measure \(90^\circ\).
(iv) Kanchanjunga is in India and Everest is in Nepal.
Answer:
(i) Let \(p\) : \(\Delta\text{ABC}\) is not equilateral.
\(q\) : It is not equiangular.
Then the symbolic form of the given statement is \(p \rightarrow q\).
Since, \(\sim(p \rightarrow q) \equiv p \wedge \sim q\), the negation of the given statement is:
‘\(\Delta\text{ABC}\) is not equilateral and it is equiangular.’

(ii) Let \(p\) : Ramesh is intelligent.
\(q\) : He is hard working.
Then the symbolic form of the given statement is \(p \wedge q\).
Since, \(\sim(p \wedge q) \equiv \sim p \vee \sim q\), the negation of the given statement is:
‘Ramesh is not intelligent or he is not hard-working.’

(iii) Let \(p\) : An angle is a right angle.
\(q\) : It is of measure \(90^\circ\).
Then the symbolic form of the given statement is \(p \leftrightarrow q\).
Since, \(\sim(p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (q \wedge \sim p)\), the negation of the given statement is:
‘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.’

(iv) Let \(p\) : Kanchenjunga is in India.
\(q\) : Everest is in Nepal.
Then the symbolic form of the given statement is \(p \wedge q\).
Since, \(\sim(p \wedge q) \equiv \sim p \vee \sim q\), the negation of the given statement is:
‘Kanchenjunga is not in India or Everest is not in Nepal.’
In simple words: To find the negation (opposite) of logical statements, we use standard rules: the negation of "if p then q" is "p and not q"; the negation of "p and q" is "not p or not q"; and the negation of "p if and only if q" is "p and not q, or q and not p". This helps us write the exact logical opposite of any compound statement.

🎯 Exam Tip: Always define the simple statements \(p\) and \(q\) clearly before writing the symbolic form. This ensures you apply De Morgan's laws or conditional negation rules correctly to get full marks.

Question (v). If \( x \in A \cap B \), then \( x \in A \) and \( x \in B \).
Answer: Let \( p : x \in A \cap B \), \( q : x \in A \), \( r : x \in B \).
Then the symbolic form of the given statement is \( p \rightarrow (q \wedge r) \).
Since, \( \sim(p \rightarrow q) \equiv p \wedge \sim q \) and \( \sim(p \wedge q) \equiv \sim p \vee \sim q \),
the negation of the given statement is:
'\( x \in A \cap B \) and \( x \notin A \) or \( x \notin B \)'.
In simple words: To find the opposite of the statement, we show that \( x \) can be in the intersection of A and B, but not in A or not in B.

🎯 Exam Tip: Remember that the negation of an implication \( p \rightarrow q \) is \( p \wedge \sim q \). Use De Morgan's laws to negate the conjunction inside.

Question 11. Construct the truth table for each of the following statement patterns:
(i) \( (p \wedge \sim q) \leftrightarrow (q \rightarrow p) \)
(ii) \( (\sim p \vee q) \wedge (\sim p \wedge \sim q) \)
(iii) \( (p \wedge r) \rightarrow (p \vee \sim q) \)
Answer:

(i) \( (p \wedge \sim q) \leftrightarrow (q \rightarrow p) \)

(ii) \( (\sim p \vee q) \wedge (\sim p \wedge \sim q) \)

(iii) \( (p \wedge r) \rightarrow (p \vee \sim q) \)

In simple words: A truth table lists all possible true/false combinations for the statements to find the final truth value of the entire pattern.

🎯 Exam Tip: Double-check each column step-by-step, especially implications (\( \rightarrow \)), which are only false when True leads to False.

Question 1. Construct the truth table for each of the following statement patterns:
(iii) \( (p \wedge r) \rightarrow (p \vee \sim q) \)
(iv) \( (p \vee r) \rightarrow \sim(q \wedge r) \)
(v) \( (p \vee \sim q) \rightarrow (r \wedge p) \)
Answer:
(iii) \( (p \wedge r) \rightarrow (p \vee \sim q) \)

(iv) \( (p \vee r) \rightarrow \sim(q \wedge r) \)

(v) \( (p \vee \sim q) \rightarrow (r \wedge p) \)

In simple words: A truth table shows us whether a logical statement is true or false for every possible scenario of its individual parts. By breaking the statement down into smaller columns, we can easily find the final truth values step-by-step.

🎯 Exam Tip: When constructing truth tables with three variables (p, q, r), always write the 8 standard combinations systematically (4 T's then 4 F's for p, alternating pairs for q, and alternating single T/F for r) to avoid missing any cases.

Truth Table for \( (p \vee \sim q) \rightarrow (r \wedge p) \):

Question 12. What is a tautology? What is a contradiction? Show that the negation of a tautology is a contradiction and the negation of a contradiction is a tautology.
Answer:
Tautology: A statement pattern that has all the entries in the last column of its truth table as T is called a tautology.
For example, let us consider the statement pattern \( p \vee \sim p \):

In the above truth table for the statement \( p \vee \sim p \), we observe that all the entries in the last column are T. Hence, the statement \( p \vee \sim p \) is a tautology.

Contradiction: A statement pattern that has all the entries in the last column of its truth table as F is called a contradiction.
For example, let us consider the statement pattern \( p \wedge \sim p \):

In the above truth table for the statement \( p \wedge \sim p \), we observe that all the entries in the last column are F. Hence, the statement \( p \wedge \sim p \) is a contradiction.

Proof of Negation Properties:
1. Let \( T \) be a tautology. By definition, the truth value of \( T \) is always T (True) for all possible truth values of its components. The negation of \( T \), denoted by \( \sim T \), will therefore always have the truth value F (False). Since all entries in its truth table are F, \( \sim T \) is a contradiction. Thus, the negation of a tautology is a contradiction.
2. Let \( C \) be a contradiction. By definition, the truth value of \( C \) is always F (False) for all possible truth values of its components. The negation of \( C \), denoted by \( \sim C \), will therefore always have the truth value T (True). Since all entries in its truth table are T, \( \sim C \) is a tautology. Thus, the negation of a contradiction is a tautology. Understanding these fundamental properties helps in simplifying complex logical expressions in mathematical reasoning.
In simple words: A tautology is a statement that is always true, like saying 'it will rain today or it will not rain'. A contradiction is always false, like saying 'it is raining and it is not raining'. If you negate (opposite) a statement that is always true, it becomes always false, and vice versa.

🎯 Exam Tip: When defining tautology and contradiction, always draw the simple truth tables for \( p \vee \sim p \) and \( p \wedge \sim p \) to secure full marks.

Truth Table for \( p \wedge \sim p \)

In the above truth table for the statement \( p \wedge \sim p \), we observe that all the entries in the last column are F. Hence, the statement \( p \wedge \sim p \) is a contradiction.

To show that the negation of a tautology is a contradiction and vice versa:
A tautology is true on every row of its truth table. Since, \( \sim T = F \) and \( \sim F = T \), when we negate a tautology, the resulting statement is false on every row of its table. i.e. the negation of tautology is a contradiction. Similarly, the negation of a contradiction is a tautology.

Question 13. Determine whether the following statement patterns is a tautology or a contradiction or a contingency:
(i) \( [(p \wedge q) \vee (\sim p)] \vee [p \wedge (\sim q)] \)
(ii) \( [(\sim p \wedge q) \wedge (q \wedge r)] \vee (\sim q) \)
Answer:
(i) \( [(p \wedge q) \vee (\sim p)] \vee [p \wedge (\sim q)] \)
Let us construct the truth table for the given statement pattern:

All the entries in the last column of the above truth table are T. Therefore, the statement pattern \( [(p \wedge q) \vee (\sim p)] \vee [p \wedge (\sim q)] \) is a tautology.

(ii) \( [(\sim p \wedge q) \wedge (q \wedge r)] \vee (\sim q) \)
Let us construct the truth table for the given statement pattern:

The entries in the last column of the above truth table are a combination of T and F. Therefore, the statement pattern \( [(\sim p \wedge q) \wedge (q \wedge r)] \vee (\sim q) \) is a contingency.
In simple words: A statement is a tautology if it is always true, a contradiction if it is always false, and a contingency if it can be either true or false depending on the inputs. Here, the first statement is always true, while the second statement changes based on the values of p, q, and r.

🎯 Exam Tip: When constructing truth tables with three variables, ensure you have exactly 8 rows to cover all possible combinations of T and F, and double-check each logical connective step-by-step.

Question (ii) Classify the statement pattern \( [(\sim p \wedge q) \wedge (q \wedge r)] \vee (\sim q) \) as a tautology, contradiction or contingency.
Answer: Let us construct the truth table for the given statement pattern:

The entries in the last column of the above truth table are neither all T nor all F. This indicates that the truth value of the statement varies depending on the truth values of its components.
\( \implies [(\sim p \wedge q) \wedge (q \wedge r)] \vee (\sim q) \) is a contingency.
In simple words: Since the final column contains a mix of both True (T) and False (F) values, the statement is neither always true nor always false, making it a contingency.

🎯 Exam Tip: Double-check each row's conjunction and disjunction operations carefully, as a single error in truth values can incorrectly classify the entire statement pattern.

Question (iii) Classify the statement pattern \( [ \sim(p \vee q) \rightarrow p ] \leftrightarrow [ (\sim p) \wedge (\sim q) ] \) as a tautology, contradiction or contingency.
Answer: Let us construct the truth table for the given statement pattern:

All the entries in the last column of the above truth table are F. This uniform result demonstrates that the statement is logically false under every possible scenario.
\( \implies [\sim(p \vee q) \rightarrow p] \leftrightarrow [(\sim p) \wedge (\sim q)] \) is a contradiction.
In simple words: Since every single entry in the final column is False (F), the statement pattern is always false under all conditions, which means it is a contradiction.

🎯 Exam Tip: Remember that a conditional statement \( A \rightarrow B \) is only False when \( A \) is True and \( B \) is False. Keep this rule in mind to avoid mistakes in intermediate columns.

Question (iv) Classify the statement pattern \( [ \sim(p \wedge q) \rightarrow p ] \leftrightarrow [ (\sim p) \wedge (\sim q) ] \) as a tautology, contradiction or contingency.
Answer: Let us construct the truth table for the given statement pattern:

The entries in the last column of the above truth table are neither all T nor all F. This mixture of truth values shows that the statement's validity is conditional upon the inputs.
\( \implies [\sim(p \wedge q) \rightarrow p] \leftrightarrow [(\sim p) \wedge (\sim q)] \) is a contingency.
In simple words: Because the final column contains both True (T) and False (F) values, the statement pattern is a contingency since its truth value depends on the individual statements.

🎯 Exam Tip: When constructing truth tables, clearly number your columns so you can easily refer to them when performing operations like biconditional (\( \leftrightarrow \)) or conditional (\( \rightarrow \)) steps.

Question 13. (v) Using the truth table, classify the statement pattern: \( [p \rightarrow (\sim q \vee r)] \leftrightarrow \sim[p \rightarrow (q \rightarrow r)] \)
Answer:

All the entries in the last column of the above truth table are F. This means that the statement is false under all possible truth value assignments of its components.
\( \therefore [p \rightarrow (\sim q \vee r)] \leftrightarrow \sim[p \rightarrow (q \rightarrow r)] \) is a contradiction.
In simple words: A contradiction is a statement that is always false, no matter what the individual parts are. Since the final column of our table has only 'F' values, this statement is a contradiction.

🎯 Exam Tip: Double-check your truth values for conditional statements (\( \rightarrow \)), remembering that they are only false when a true statement leads to a false one.

Question 14. Using the truth table, prove the following logical equivalences:
(i) \( p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r) \)
Answer:

The entries in columns 5 and 8 are identical. This confirms that the distributive law of conjunction over disjunction holds true.
\( \therefore p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r) \)
In simple words: To prove two statements are equivalent, we show that their columns in the truth table have the exact same T and F values in every row. Since column 5 and column 8 match perfectly, the two statements are logically equivalent.

🎯 Exam Tip: Always number the columns in your truth table and explicitly state which columns are identical to conclude your proof clearly.

Question (ii). Prove that \( [ \sim(p \vee q) \vee (p \vee q) ] \wedge r \equiv r \) using a truth table.
Answer: To prove the equivalence, we construct the truth table for both sides of the statement:

The entries in columns 3 and 7 are identical. This truth table clearly shows that the truth values of both expressions are identical under all possible logical interpretations.
\( \therefore [ \sim(p \vee q) \vee (p \vee q) ] \wedge r \equiv r \)
In simple words: We can see that the column for the final statement and the column for 'r' have the exact same True and False values in every row. This proves that both statements mean the exact same thing.

🎯 Exam Tip: Always number your columns in truth tables so you can easily refer to them when writing your final concluding statement.

Question (iii). Prove that \( p \wedge (\sim p \vee q) \equiv p \wedge q \) using a truth table.
Answer: To prove the equivalence, we construct the truth table for both sides of the statement:

The entries in columns 5 and 6 are identical. This logical equivalence demonstrates how the distributive property simplifies complex logical conjunctions.
\( \therefore p \wedge (\sim p \vee q) \equiv p \wedge q \)
In simple words: By comparing columns 5 and 6, we find that they have identical truth values in every single row. This shows that the two logical expressions are completely equivalent.

🎯 Exam Tip: Double-check the negation column (\( \sim p \)) first, as a small mistake there will carry over and make the entire table incorrect.

Question (iv). Prove that \( p \leftrightarrow q \equiv \sim(p \wedge \sim q) \wedge \sim(q \wedge \sim p) \) using a truth table.
Answer: To prove the equivalence, we construct the truth table for both sides of the statement:

The entries in columns 9 and 10 are identical. This equivalence shows that a biconditional statement is true if and only if both conditional directions are simultaneously satisfied.
\( \therefore p \leftrightarrow q \equiv \sim(p \wedge \sim q) \wedge \sim(q \wedge \sim p) \)
In simple words: We can see that the column for the double arrow (if and only if) and the column for the long combined statement have the exact same True and False values. This proves they are logically identical.

🎯 Exam Tip: For complex statements with many columns, write down each sub-expression step-by-step to avoid making calculation errors.

Question (v) Prove that \( \sim p \wedge q \equiv (p \vee q) \wedge \sim p \) using a truth table.
Answer:
To prove the equivalence, we construct the truth table. This systematic approach allows us to compare the truth values of both statements under all possible truth value assignments of their components.

The entries in columns 4 and 6 are identical.
\( \therefore \sim p \wedge q \equiv (p \vee q) \wedge \sim p \)
In simple words: By comparing the truth values for both statements under all possible scenarios, we see their final columns match exactly, proving they mean the same thing.

🎯 Exam Tip: Clearly number your columns in the truth table and explicitly state which columns have identical entries to secure full marks.

Question 15. Write the converse, inverse, contrapositive of the following statements:
(i) If 2 + 5 = 10, then 4 + 10 = 20.
Answer:
Let \( p \): 2 + 5 = 10
\( q \): 4 + 10 = 20
Then the symbolic form of the given statement is \( p \rightarrow q \). This representation helps us systematically apply logical rules to find the related statements.
Converse: \( q \rightarrow p \) is the converse of \( p \rightarrow q \)
i.e. If 4 + 10 = 20, then 2 + 5 = 10.
Inverse: \( \sim p \rightarrow \sim q \) is the inverse of \( p \rightarrow q \)
i.e. If 2 + 5 \( \neq \) 10, then 4 + 10 \( \neq \) 20.
Cotrapositive: \( \sim q \rightarrow \sim p \) is the contrapositive of \( p \rightarrow q \),
i.e. If 4 + 10 \( \neq \) 20, then 2 + 5 \( \neq \) 10.
In simple words: For any conditional statement "If P then Q", its converse swaps them ("If Q then P"), its inverse negates both ("If not P then not Q"), and its contrapositive does both ("If not Q then not P").

🎯 Exam Tip: Remember that a conditional statement is always logically equivalent to its contrapositive, which is a great way to double-check your logic.

Question 15. Write the converse, inverse, and contrapositive of the following statements:
(ii) If a man is a bachelor, then he is happy.
(iii) If I do not work hard, then I do not prosper.
Answer:
(ii) Let \( p \) : A man is a bachelor.
\( q \) : He is happy.
Then the symbolic form of the given statement is \( p \rightarrow q \).
Converse: \( q \rightarrow p \) is the converse of \( p \rightarrow q \)
i.e. If a man is happy, then he is a bachelor.
Inverse: \( \sim p \rightarrow \sim q \) is the inverse of \( p \rightarrow q \)
i.e. If a man is not a bachelor, then he is not happy.
Contrapositive: \( \sim q \rightarrow \sim p \) is the contrapositive of \( p \rightarrow q \)
i.e., If a man is not happy, then he is not a bachelor.

(iii) Let \( p \) : I do not work hard.
\( q \) : I do not prosper.
Then the symbolic form of the given statement is \( p \rightarrow q \).
Converse: \( q \rightarrow p \) is the converse of \( p \rightarrow q \)
i.e. If I do not prosper, then I do not work hard.
Inverse: \( \sim p \rightarrow \sim q \) is the inverse of \( p \rightarrow q \)
i.e. If I work hard, then I prosper.
Contrapositive: \( \sim q \rightarrow \sim p \) is the contrapositive of \( p \rightarrow q \)
i.e. If I prosper, then I work hard.
In simple words: The converse swaps the two statements, the inverse negates both statements, and the contrapositive both swaps and negates them.

🎯 Exam Tip: Always define the simple statements \( p \) and \( q \) clearly before writing their symbolic forms to avoid any confusion during translation.

Question 16. State the dual of each of the following statements by applying the principle of duality:
(i) \( (p \wedge \sim q) \vee (\sim p \wedge q) \equiv (p \vee q) \wedge \sim(p \wedge q) \)
(ii) \( p \vee (q \vee r) \equiv \sim[(p \wedge q) \vee (r \vee s)] \)
(iii) 2 is an even number or 9 is a perfect square.
Answer:
The duals are given by:
(i) \( (p \vee \sim q) \wedge (\sim p \vee q) \equiv (p \wedge q) \vee \sim(p \vee q) \)
(ii) \( p \wedge (q \wedge r) \equiv \sim[(p \vee q) \wedge (r \wedge s)] \)
(iii) 2 is an even number and 9 is a perfect square.
In simple words: To find the dual of a statement, replace the 'or' (\( \vee \)) operator with 'and' (\( \wedge \)), and replace 'and' (\( \wedge \)) with 'or' (\( \vee \)) without changing any negations.

🎯 Exam Tip: Remember that the negation symbol \( \sim \) remains completely unchanged when writing the dual of a statement.

Question 17. Rewrite the following statements without using the connective ‘If … then’:
(i) If a quadrilateral is a rhombus, then it is not a square.
(ii) If \( 10 - 3 = 7 \), then \( 10 \times 3 \neq 30 \).
(iii) If it rains, then the principal declares a holiday.
Answer:
Since, \( p \rightarrow q \equiv \sim p \vee q \), the given statements can be written as:
(i) A quadrilateral is not a rhombus or it is not a square.
(ii) \( 10 - 3 \neq 7 \) or \( 10 \times 3 \neq 30 \).
(iii) It does not rain or the principal declares a holiday. This logical equivalence helps simplify conditional statements into disjunctions.
In simple words: To remove 'If... then' (which is \( p \rightarrow q \)), we can rewrite it as 'not p or q' (which is \( \sim p \vee q \)). This keeps the exact same logical meaning but uses 'or' instead of 'if'.

🎯 Exam Tip: Remember the standard equivalence \( p \rightarrow q \equiv \sim p \vee q \). Negate the antecedent (first part), change 'if-then' to 'or', and keep the consequent (second part) as it is.

Question 18. Write the dual of each of the following:
(i) \( (\sim p \wedge q) \vee (p \wedge \sim q) \vee (\sim p \wedge \sim q) \)
(ii) \( (p \wedge q) \wedge r \equiv p \wedge (q \wedge r) \)
(iii) \( p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \)
(iv) \( \sim(p \vee q) \equiv \sim p \wedge \sim q \)
Answer:
The duals are given by:
(i) \( (\sim p \vee q) \wedge (p \vee \sim q) \wedge (\sim p \vee \sim q) \)
(ii) \( (p \vee q) \vee r \equiv p \vee (q \vee r) \)
(iii) \( p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r) \)
(iv) \( \sim(p \wedge q) \equiv \sim p \vee \sim q \)
Finding the dual of a statement is a fundamental operation in mathematical logic that interchanges conjunctions and disjunctions.
In simple words: To find the dual of a logical statement, you simply swap every 'and' (\( \wedge \)) with 'or' (\( \vee \)), and every 'or' (\( \vee \)) with 'and' (\( \wedge \)). The negations (\( \sim \)) remain exactly the same.

🎯 Exam Tip: When writing duals, only replace \( \wedge \) with \( \vee \), \( \vee \) with \( \wedge \), \( T \) with \( F \), and \( F \) with \( T \). Do not change any negation symbols (\( \sim \)).

Question 19. Consider the following statements:
(i) If D is a dog, then D is very good.
(ii) If D is very good, then D is a 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.
Answer:
Let \( p \): D is a dog and \( q \): D is very good.
Then the given statements in the symbolic form are:
(i) \( p \rightarrow q \)
(ii) \( q \rightarrow p \)
(iii) \( \sim q \rightarrow \sim p \)
(iv) \( \sim p \rightarrow \sim q \)
Since a conditional statement is logically equivalent to its contrapositive, we have \( p \rightarrow q \equiv \sim q \rightarrow \sim p \) and \( q \rightarrow p \equiv \sim p \rightarrow \sim q \). Therefore, statements (i) and (iii) have the same meaning, and statements (ii) and (iv) have the same meaning. This demonstrates how a conditional statement and its contrapositive always share the same truth value.
In simple words: A statement 'If A, then B' always means the exact same thing as 'If not B, then not A' (its contrapositive). Therefore, statements (i) and (iii) match, and statements (ii) and (iv) match.

🎯 Exam Tip: Always write down the symbolic form of each statement first. It makes it very easy to identify equivalent pairs like conditional and contrapositive, or converse and inverse.

Question 20. Express the truth of each of the following statements by Venn diagrams:
(i) All men are mortal.
(ii) Some persons are not politicians.
Answer:
(i) Let \( U \) be the universal set of all human beings.
Let \( A \) be the set of all men.
Let \( B \) be the set of all mortals.
Since all men are mortal, every member of set \( A \) is also a member of set \( B \), which means \( A \subseteq B \). This relationship is represented by drawing circle \( A \) entirely inside circle \( B \).

(ii) Let \( U \) be the universal set of all human beings.
Let \( A \) be the set of all persons.
Let \( B \) be the set of all politicians.
The statement "Some persons are not politicians" means there are some individuals in set \( A \) who do not belong to set \( B \). This indicates that the region \( A - B \) is non-empty, which is represented by shading the portion of circle \( A \) that lies outside circle \( B \). This visual representation clearly demonstrates how logical statements can be translated into set relationships.

In simple words: We use Venn diagrams to show how different groups of things relate to each other. For the first part, since all men are mortal, the circle for 'men' sits completely inside the bigger circle for 'mortals'. For the second part, because some people are not politicians, we shade the part of the 'persons' circle that does not overlap with the 'politicians' circle.

🎯 Exam Tip: Always define the universal set \( U \) and clearly label each set with capital letters. Shading the correct region or showing proper containment is crucial for scoring full marks in Venn diagram questions.

p>Question (iii) Some members of the present Indian cricket are not committed.
Answer:
Let \( U \) : set of all human beings
\( X \) : set of all members of present Indian cricket
\( Y \) : set of all committed members of the present Indian cricket.
Then the Venn diagram represents the truth of the given statement is as below:

\( X - Y \neq \phi \)
In simple words: The shaded ring represents the members of the Indian cricket team who are not committed. Since this shaded region is not empty, it shows some members are indeed not committed.

🎯 Exam Tip: Clearly define your sets \( U \), \( X \), and \( Y \) before drawing the Venn diagram, and always write the set relation (like \( X - Y \neq \phi \)) below the diagram to secure full marks.

Question (iv) No child is an adult.
Answer:
Let \( U \) : set of all human beings
\( C \) : set of all children
\( A \) : set of all adults.
Then the Venn diagram represents the truth of the given statement is as below:

\( C \cap A = \phi \)
In simple words: Since children and adults are completely different groups with no one belonging to both, their circles are drawn separately without overlapping.

🎯 Exam Tip: When representing "No A is B", always draw two completely separate (disjoint) circles inside the universal set rectangle and state that their intersection is empty (\( \phi \)).

Question 21. If \( A = \{2, 3, 4, 5, 6, 7, 8\} \), determine the truth value of each of the following

Question. Determine the truth value of the following statements:
(i) \( \exists x \in A \), such that \( 3x + 2 > 9 \).
(ii) \( \forall x \in A, x^2 < 18 \).
(iii) \( \exists x \in A \), such that \( x + 3 < 11 \).
(iv) \( \forall x \in A, x^2 + 2 \ge 5 \).
Answer:
(i) Clearly \( x = 3, 4, 5, 6, 7, 8 \in A \) satisfy \( 3x + 2 > 9 \). So, the given statement is true, hence its truth value is T.
(ii) \( x = 5, 6, 7, 8 \in A \) do not satisfy \( x^2 < 18 \). So the given statement is false, hence its truth value is F.
(iii) Clearly \( x = 2, 3, 4, 5, 6, 7 \in A \) which satisfy \( x + 3 < 11 \). So, the given statement is True, hence its truth value is T.
(iv) \( x^2 + 2 \ge 5 \) for all \( x \in A \). So, the given statement is true, hence its truth value is T. This shows how existential and universal quantifiers behave differently under the same set.
In simple words: For 'there exists' (\( \exists \)), we only need one number from set A to make the statement true. For 'for all' (\( \forall \)), every single number in set A must satisfy the condition.

🎯 Exam Tip: When dealing with quantifiers, remember that \( \exists \) (existential) requires only one true case, while \( \forall \) (universal) requires all cases to be true to make the statement true.

Question 22. Write the negations of the following statements:
(i) 7 is a prime number and the Taj Mahal is in Agra.
(ii) 10 > 5 and 3 < 8.
Answer:
(i) Let \( p \): 7 is a prime number.
\( q \): Taj Mahal is in Agra.
Then the symbolic form of the given statement is \( p \wedge q \).
Since, \( \sim(p \wedge q) \equiv \sim p \vee \sim q \), the negation of the given statement is: ‘7 is not a prime number or Taj Mahal is not in Agra.’
(ii) Let \( p \): \( 10 > 5 \).
\( q \): \( 3 < 8 \).
Then the symbolic form of the given statement is \( p \wedge q \).
Since, \( \sim(p \wedge q) \equiv \sim p \vee \sim q \), the negation of the given statement is: ‘\( 10 \le 5 \) or \( 3 \ge 8 \)’. Applying De Morgan's laws helps us systematically write these logical negations without changing the core meaning.
In simple words: To find the negation of an "and" statement, we negate both parts and change "and" to "or" using De Morgan's Laws.

🎯 Exam Tip: Remember to change 'and' (\( \wedge \)) to 'or' (\( \vee \)) when negating a conjunction, and always define your component statements \( p \) and \( q \) clearly.

Question. Write the negation of the following statements:
(iii) I will have tea or coffee.
(iv) \( \forall n \in N, n + 3 > 9 \)
(v) \( \exists x \in A \), such that \( x + 5 < 11 \)
Answer:
(iii) The negation of the given statement is: 'I will not have tea and coffee.'
(iv) The negation of the given statement is: '\( \exists n \in N \), such that \( n + 3 \ngtr 9 \)' or '\( \exists n \in N \), such that \( n + 3 \le 9 \)'
(v) The negation of the given statement is: '\( \forall x \in A, x + 5 \nless 11 \)' or '\( \forall x \in A, x + 5 \ge 11 \)'
In simple words: To write the negation of a statement, we change "or" to "and", "for all" (\( \forall \)) to "there exists" (\( \exists \)), and reverse the inequality signs to their exact opposites.

🎯 Exam Tip: When negating quantified statements, always change the quantifier (\( \forall \leftrightarrow \exists \)) and negate the statement that follows it to secure full marks.