Samacheer Kalvi Class 12 Maths Solutions Chapter 12 Discrete Mathematics Exercise 12.3

Choose the most suitable answer from the given four alternatives:

 

Question 1. A binary operation on a set S is a function from
(a) S → S
(b) (S x S) → S
(c) S → (S x S)
Answer: (b) (S x S) → S
In simple words: A binary operation means you take two elements from a set, combine them, and the result is also in that same set. The notation (S x S) → S shows this perfectly: it takes two elements from S (written as S x S) and gives one result back into S.

🎯 Exam Tip: Remember the definition of a binary operation: it's a function that maps elements from the Cartesian product of a set with itself back into the original set, ensuring closure.

 

Question 2. Subtraction is not a binary operation in
(a) R
(b) Z
(c) N
(d) Q
Answer: (c) N
In simple words: Subtraction does not always work in the set of natural numbers (N). For example, if you take 1 and 2, both are natural numbers. But 1 - 2 equals -1, and -1 is not a natural number. This means the result is not always in the original set N.

🎯 Exam Tip: To check if an operation is binary for a given set, test a few examples, especially edge cases. If even one example fails to produce a result within the set, it's not a binary operation.

 

Question 3. Which one of the following is a binary operation on N?
(a) Subtraction
(b) Multiplication
(c) Division
(d) All of the options
Answer: (b) Multiplication
In simple words: When you multiply any two natural numbers, the answer is always another natural number. For instance, 2 x 3 = 6, and 6 is a natural number. This works every time, making multiplication a binary operation for natural numbers.

🎯 Exam Tip: Operations like subtraction (2 - 3 = -1, not in N) and division (2 / 3, not in N) do not satisfy the closure property for natural numbers.

 

Question 4. In the set R of real numbers '*' is defined as follows. Which one of the following is not a binary operation on R ?
(b) \( a * b = \max (a, b) \)
(c) \( a * b = a \)
(d) \( a * b = a^b \)
Answer: (d) \( a * b = a^b \)
In simple words: The operation \( a * b = a^b \) is not always defined for real numbers in a way that gives another real number. For example, if you take -2 and 1/2, both are real numbers. But \( (-2)^{1/2} \) (which is the square root of -2) is not a real number; it's a complex number.

🎯 Exam Tip: Pay close attention to the domain and codomain of the operation. For real numbers, taking even roots of negative numbers results in non-real (complex) numbers, violating closure for real numbers.

 

Question 5. The operation * defined by \( a * b = \frac { ab }{ 7 } \) is not a binary operation on
(a) Q+
(b) Z
(c) R
(d) C
Answer: (b) Z
In simple words: For the set of integers (Z), if you perform the operation \( a * b = \frac { ab }{ 7 } \), the result might not be an integer. For instance, if a = 1 and b = 1, then \( 1 * 1 = \frac{1 \times 1}{7} = \frac{1}{7} \), which is not an integer. This means the operation is not closed on Z.

🎯 Exam Tip: Closure is key. If an operation on elements of a set can produce a result outside that set, it is not a binary operation on that set. Integers need division by 7 to yield an integer.

 

Question 6. In the set Q, define \( a \bigodot b = a + b + ab \). For what value of y, \( 3 \bigodot (y \bigodot 5) = 7 \)?
(a) \( y = \frac { 2 }{ 3 } \)
(b) \( y = \frac { -2 }{ 3 } \)
(c) \( y = \frac { -3 }{ 2 } \)
(d) \( y = 4 \)
Answer: (b) \( y = \frac { -2 }{ 3 } \)
In simple words: We are given a new way to combine numbers, \( a \bigodot b = a + b + ab \). We need to solve for 'y' by working through the operation step-by-step from inside the parentheses outwards until the equation \( 3 \bigodot (y \bigodot 5) = 7 \) is satisfied. First, we find \( y \bigodot 5 \), then use that result with 3.
Given \( a \bigodot b = a + b + ab \)
We have \( 3 \bigodot (y \bigodot 5) = 7 \)
First, calculate \( y \bigodot 5 \):
\( y \bigodot 5 = y + 5 + y(5) = y + 5 + 5y = 6y + 5 \)
Now substitute this back into the main equation:
\( 3 \bigodot (6y + 5) = 7 \)
Using the definition again for \( 3 \bigodot (6y + 5) \):
\( 3 + (6y + 5) + 3(6y + 5) = 7 \)
\( 3 + 6y + 5 + 18y + 15 = 7 \)
Combine like terms:
\( 24y + 23 = 7 \)
Subtract 23 from both sides:
\( 24y = 7 - 23 \)
\( 24y = -16 \)
Divide by 24:
\( y = \frac { -16 }{ 24 } \)
Simplify the fraction:
\( y = \frac { -2 }{ 3 } \)

🎯 Exam Tip: Always solve operations inside parentheses first. Be careful with algebraic steps, especially combining like terms and handling negative signs.

 

Question 7. If \( a * b = \sqrt {a^2+b^2 } \) on the real numbers then * is
(a) Commutative but not associative
(b) Associative but not commutative
(c) Both commutative and associative
(d) Neither commutative nor associative
Answer: (c) Both commutative and associative
In simple words: The operation `*` is both commutative and associative. Commutative means the order doesn't matter (a*b = b*a). Associative means how you group the numbers with parentheses doesn't change the result ((a*b)*c = a*(b*c)). This particular square root operation satisfies both properties.
To check for commutativity:
\( a * b = \sqrt{a^2 + b^2} \)
\( b * a = \sqrt{b^2 + a^2} \)
Since \( a^2 + b^2 = b^2 + a^2 \), we have \( \sqrt{a^2 + b^2} = \sqrt{b^2 + a^2} \).
\( \implies a * b = b * a \)
Thus, the operation * is commutative.

To check for associativity:
We need to check if \( (a * b) * c = a * (b * c) \).
First, calculate \( (a * b) * c \):
\( (a * b) * c = \sqrt{a^2 + b^2} * c \)
\( = \sqrt{(\sqrt{a^2 + b^2})^2 + c^2} \)
\( = \sqrt{a^2 + b^2 + c^2} \)

Next, calculate \( a * (b * c) \):
\( a * (b * c) = a * \sqrt{b^2 + c^2} \)
\( = \sqrt{a^2 + (\sqrt{b^2 + c^2})^2} \)
\( = \sqrt{a^2 + b^2 + c^2} \)

Since \( (a * b) * c = \sqrt{a^2 + b^2 + c^2} \) and \( a * (b * c) = \sqrt{a^2 + b^2 + c^2} \),
\( \implies (a * b) * c = a * (b * c) \)
Thus, the operation * is associative.

🎯 Exam Tip: Always write out the definitions for commutativity and associativity and substitute the given operation to prove or disprove each property systematically. Simplify both sides to compare them.

 

Question 8. Which one of the following statements has the truth value T?
(a) sin x is an even function.
(b) Every square matrix is non-singular
(c) The product of complex number and its conjugate is purely imaginary
(d) \( \sqrt{5} \) is an irrational number
Answer: (d) \( \sqrt{5} \) is an irrational number
In simple words: The square root of 5 cannot be written as a simple fraction, meaning it's an irrational number. This statement is true. Other options are false: sin x is an odd function, not all square matrices are non-singular (some have determinant 0), and the product of a complex number and its conjugate is always a real number, not purely imaginary.

🎯 Exam Tip: Understand fundamental definitions in mathematics, such as properties of functions, matrices, complex numbers, and number systems, to quickly identify true and false statements.

 

Question 9. Which one of the following statements has truth value F?
(a) Chennai is in India or \( \sqrt{2} \) is an integer.
(b) Chennai is in China or \( \sqrt{2} \) is an irrational number.
(c) Chennai is in China or \( \sqrt{2} \) is an integer.
(d) Chennai is in China or \( \sqrt{2} \) is an irrational number.
Answer: (c) Chennai is in China or \( \sqrt{2} \) is an integer.
In simple words: This statement is false because both parts of it are false. Chennai is not in China, and the square root of 2 is not an integer. Since an 'or' statement is only true if at least one part is true, and both parts here are false, the whole statement is false.
Let p be "Chennai is in China" (False).
Let q be "\( \sqrt{2} \) is an integer" (False).
The compound statement is \( p \lor q \).
\( F \lor F = F \). Thus, statement (c) has a truth value of False.

🎯 Exam Tip: For 'or' statements (\( p \lor q \)), remember that the statement is true if at least one of p or q is true. It is false only if both p and q are false.

 

Question 10. If a compound statement involves 3 simple statements, then the number of rows in the truth table is ..........
(a) 9
(b) 8
(c) 6
(d) 3
Answer: (b) 8
In simple words: For a truth table, if you have 'n' simple statements, the total number of rows you need is \( 2^n \). Since there are 3 simple statements, you need \( 2^3 \), which is 8 rows. This ensures all possible combinations of truth values for the statements are covered.

🎯 Exam Tip: The number of rows in a truth table is always \( 2^n \), where n is the number of distinct simple statements involved. This formula is critical for constructing truth tables correctly.

 

Question 11. Which one is the inverse of the statement \( (p \lor q) \rightarrow (p \land q) \)?
(a) \( (p \land q) \rightarrow (p \lor q) \)
(b) \( \neg(p \lor q) \rightarrow (p \land q) \)
(c) \( (\neg P \lor \neg q) \rightarrow (\neg p \land \neg q) \)
(d) \( (\neg p \land \neg q) \rightarrow (\neg p \lor \neg q) \)
Answer: (d) \( (\neg p \land \neg q) \rightarrow (\neg p \lor \neg q) \)
In simple words: The inverse of a conditional statement \( A \rightarrow B \) is \( \neg A \rightarrow \neg B \). We apply this rule to the given statement \( (p \lor q) \rightarrow (p \land q) \). We find the negation of the first part, which is \( \neg(p \lor q) \), and the negation of the second part, which is \( \neg(p \land q) \). Then, using De Morgan's laws, \( \neg(p \lor q) \) becomes \( \neg p \land \neg q \), and \( \neg(p \land q) \) becomes \( \neg p \lor \neg q \).
Given statement: \( (p \lor q) \rightarrow (p \land q) \)
The inverse of a conditional statement \( A \rightarrow B \) is \( \neg A \rightarrow \neg B \).
Here, \( A = (p \lor q) \) and \( B = (p \land q) \).
So, \( \neg A = \neg(p \lor q) \) and \( \neg B = \neg(p \land q) \).
Using De Morgan's Laws:
\( \neg(p \lor q) \equiv \neg p \land \neg q \)
\( \neg(p \land q) \equiv \neg p \lor \neg q \)
Therefore, the inverse statement is \( (\neg p \land \neg q) \rightarrow (\neg p \lor \neg q) \).

🎯 Exam Tip: Always remember the definitions for converse, inverse, and contrapositive. For a statement \( P \rightarrow Q \): converse is \( Q \rightarrow P \), inverse is \( \neg P \rightarrow \neg Q \), and contrapositive is \( \neg Q \rightarrow \neg P \).

 

Question 12. Which one is the contrapositive of the statement \( (p \lor q) \rightarrow r \)?
(a) \( \neg r \rightarrow (\neg p \land \neg q) \)
(b) \( \neg r \rightarrow (p \lor q) \)
(c) \( r \rightarrow (p \land q) \)
(d) \( p \rightarrow (q \lor r) \)
Answer: (a) \( \neg r \rightarrow (\neg p \land \neg q) \)
In simple words: The contrapositive of a statement \( A \rightarrow B \) is \( \neg B \rightarrow \neg A \). Here, \( A \) is \( (p \lor q) \) and \( B \) is \( r \). So we need to negate \( r \) and make it the new start, and negate \( (p \lor q) \) and make it the new end. The negation of \( (p \lor q) \) is \( \neg p \land \neg q \) by De Morgan's laws.
Given statement: \( (p \lor q) \rightarrow r \)
The contrapositive of \( A \rightarrow B \) is \( \neg B \rightarrow \neg A \).
Here, \( A = (p \lor q) \) and \( B = r \).
So, \( \neg B = \neg r \) and \( \neg A = \neg(p \lor q) \).
Using De Morgan's Law, \( \neg(p \lor q) \equiv \neg p \land \neg q \).
Therefore, the contrapositive is \( \neg r \rightarrow (\neg p \land \neg q) \).

🎯 Exam Tip: The contrapositive always has the same truth value as the original conditional statement. Understanding De Morgan's laws is crucial for correctly negating compound statements.

 

Question 13. The truth table for \( (p \land q) \lor \neg q \) is given below. Which one of the following is true?
(a) T T T T
(b) T F T T
(c) T T F T
(d) T F F F
Answer: (c) T T F T
In simple words: We need to build the truth table for the statement \( (p \land q) \lor \neg q \) and find which row of truth values matches the result. First, find \( p \land q \), then \( \neg q \), and finally combine them with 'or'. The sequence of truth values we get is T, T, F, T.
Let's construct the truth table for \( (p \land q) \lor \neg q \):

The final column for \( (p \land q) \lor \neg q \) is T, T, F, T. This matches option (c).

🎯 Exam Tip: When constructing truth tables for compound statements, always work systematically by evaluating the simplest components first (like \( \neg q \), \( p \land q \)) before combining them into more complex expressions. This reduces errors.

 

Question 14. In the last column of the truth table for \( \neg(p \lor \neg q) \) the number of final outcomes of the truth value 'F' is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: We need to create a truth table for \( \neg(p \lor \neg q) \) and then count how many times 'F' appears in the final result column. First, calculate \( \neg q \), then \( p \lor \neg q \), and finally its negation. Counting the 'F's gives us the answer.
Let's construct the truth table for \( \neg(p \lor \neg q) \):

The truth values in the last column are F, F, T, F. There are 3 outcomes with the truth value 'F'.

🎯 Exam Tip: Always double-check your truth table calculations, especially when dealing with negations and multiple logical operators. A small error in one column can lead to an incorrect final count.

 

Question 15. Which one of the following is incorrect? For any two propositions p and q, we have
(a) \( \neg(p \lor q) \equiv \neg p \land \neg q \)
(b) \( \neg(p \land q) \equiv \neg p \lor \neg q \)
(c) \( \neg(p \lor q) \equiv \neg p \lor \neg q \)
(d) \( \neg(\neg p) \equiv p \)
Answer: (c) \( \neg(p \lor q) \equiv \neg p \lor \neg q \)
In simple words: According to De Morgan's laws, the negation of \( p \lor q \) should be \( \neg p \land \neg q \) (changing 'or' to 'and'). Option (c) wrongly states it as \( \neg p \lor \neg q \). This makes option (c) the incorrect statement. Options (a), (b), and (d) are all correct logical equivalences.

🎯 Exam Tip: Memorize De Morgan's Laws: \( \neg(p \land q) \equiv \neg p \lor \neg q \) and \( \neg(p \lor q) \equiv \neg p \land \neg q \). These are fundamental in propositional logic and frequently tested.

 

Question 16. Which one of the following is correct for the truth value of \( (p \land q) \rightarrow \neg p \)?
(1) (2) (3) (4)
(a) T T T T
(b) F T T T
(c) F F T T
(d) T T T F
Answer: (b) (1) F, (2) T, (3) T, (4) T
In simple words: We need to determine the truth values for the expression \( (p \land q) \rightarrow \neg p \). We build a truth table by first finding \( p \land q \), then \( \neg p \), and finally combining them with the 'if-then' (conditional) operator. The resulting sequence of truth values is F, T, T, T.
Let's construct the truth table for \( (p \land q) \rightarrow \neg p \):

The truth values in the last column are F, T, T, T. This matches option (b).

🎯 Exam Tip: Remember that a conditional statement \( A \rightarrow B \) is only false when A is true and B is false. In all other cases, it is true.

 

Question 17. The dual of \( \neg(p \lor q) \lor [p \lor (p \land \neg r)] \) is
(a) \( \neg(p \land q) \land [p \lor (p \land \neg r)] \)
(b) \( (p \land q) \land [p \lor p \lor \neg r)] \)
(c) \( \neg(p \land q) \land [p \land (p \land r)] \)
(d) \( \neg(p \land q) \land [p \land (p \lor \neg r)] \)
Answer: (d) \( \neg(p \land q) \land [p \land (p \lor \neg r)] \)
In simple words: To find the dual of a logical expression, we swap 'or' (\( \lor \)) with 'and' (\( \land \)) and 'and' (\( \land \)) with 'or' (\( \lor \)). We also swap T with F and F with T, but here there are no T or F values, so only the operators change. Negations stay as they are.
Given statement: \( \neg(p \lor q) \lor [p \lor (p \land \neg r)] \)
To find the dual, replace \( \lor \) with \( \land \) and \( \land \) with \( \lor \). Negations remain unchanged.
Dual: \( \neg(p \land q) \land [p \land (p \lor \neg r)] \)

🎯 Exam Tip: When finding the dual, remember that negations are not changed. Only the logical connectives (\( \land \), \( \lor \)) and truth values (T, F) are interchanged.

 

Question 18. The proposition \( p \land (\neg p \lor q) \) is ........
(a) a tautology
(b) a contradiction
(c) logically equivalent to \( p \land q \)
(d) logically equivalent to \( p \lor q \)
Answer: (c) logically equivalent to \( p \land q \)
In simple words: We can simplify the expression \( p \land (\neg p \lor q) \). Using the distributive law, this becomes \( (p \land \neg p) \lor (p \land q) \). Since \( p \land \neg p \) is always false (a contradiction), the expression simplifies to \( \text{F} \lor (p \land q) \), which is just \( p \land q \).
Let's simplify the given proposition \( p \land (\neg p \lor q) \).
Using the Distributive Law \( A \land (B \lor C) \equiv (A \land B) \lor (A \land C) \):
\( p \land (\neg p \lor q) \equiv (p \land \neg p) \lor (p \land q) \)
We know that \( p \land \neg p \) is always false (F).
\( \implies \text{F} \lor (p \land q) \)
Since \( \text{F} \lor X \equiv X \), the expression simplifies to:
\( \implies p \land q \)
Thus, \( p \land (\neg p \lor q) \) is logically equivalent to \( p \land q \).

🎯 Exam Tip: Look for opportunities to simplify logical expressions using known identities such as distributive laws, De Morgan's laws, and the properties of contradictions (\( p \land \neg p \equiv \text{F} \)) and tautologies (\( p \lor \neg p \equiv \text{T} \)).

 

Question 19. Determine the truth value of each of the following statements:
(a) 4 + 2 = 5 and 6 + 3 = 9
(b) 3 + 2 = 5 and 6 + 1 = 7
(c) 4 + 5 = 9 and 1 + 2 = 4
(d) 3 + 2 = 5 and 4 + 7 = 11
Answer: (a) (1) F, (2) T, (3) F, (4) T
In simple words: We need to figure out if each compound statement (made with 'and') is true or false. A statement joined by 'and' is only true if both parts are true. If even one part is false, the whole 'and' statement is false. We check each part to find the truth value.
Let's determine the truth value for each statement:
(a) \( (4 + 2 = 5) \land (6 + 3 = 9) \)
\( (F) \land (T) \implies F \)
(b) \( (3 + 2 = 5) \land (6 + 1 = 7) \)
\( (T) \land (T) \implies T \)
(c) \( (4 + 5 = 9) \land (1 + 2 = 4) \)
\( (T) \land (F) \implies F \)
(d) \( (3 + 2 = 5) \land (4 + 7 = 11) \)
\( (T) \land (T) \implies T \)
So the sequence of truth values is F, T, F, T.

🎯 Exam Tip: For compound statements linked by 'and' (\( \land \)), remember the "all or nothing" rule: the entire statement is true only if every single component statement is true. One false component makes the whole thing false.

 

Question 20. Which one of the following is not true?
(a) Negation of a statement is the statement itself.
(b) If the last column of the truth table contains only T then it is a tautology.
(c) If the last column of its truth table contains only F then it is a contradiction
(d) If p and q are any two statements then \( p \leftrightarrow q \) is a tautology.
Answer: (d) If p and q are any two statements then \( p \leftrightarrow q \) is a tautology.
In simple words: The statement \( p \leftrightarrow q \) (p if and only if q) is a tautology only when p and q always have the same truth value, which means they are logically equivalent. It's not true for *any* two statements. For example, if p is "It is raining" and q is "The sun is shining," \( p \leftrightarrow q \) would usually be false, not a tautology.

🎯 Exam Tip: Understand the precise definitions of tautology, contradiction, and logical equivalence. A tautology is always true, a contradiction is always false, and logical equivalence means two statements have identical truth tables.