CBSE Class 12 Mathematics Relations And Functions Assignment Set 09

XII - Mathematics Assignment - Relations and Functions

Q1. Let \(Z\) be the set of all integers and \(R\) be a relation on \(Z\) defined as \(R = \{(a, b) : a, b \in Z\}\) also \((a - b)\) is divisible by 5. Prove that \(R\) is an equivalence relation.
Answer: To prove that \(R\) is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.
1. Reflexive:
For any \(a \in Z\), we have \(a - a = 0\).
Since 0 is divisible by 5, \((a, a) \in R\) for all \(a \in Z\).
Therefore, \(R\) is reflexive.
2. Symmetric:
Let \((a, b) \in R\) for some \(a, b \in Z\).
This means \((a - b)\) is divisible by 5.
\(\Rightarrow a - b = 5k\) for some integer \(k\).
\(\Rightarrow b - a = 5(-k)\).
Since \(-k\) is an integer, \((b - a)\) is also divisible by 5.
\(\Rightarrow (b, a) \in R\).
Therefore, \(R\) is symmetric.
3. Transitive:
Let \((a, b) \in R\) and \((b, c) \in R\) for some \(a, b, c \in Z\).
Then \(a - b = 5k_1\) and \(b - c = 5k_2\) for some integers \(k_1, k_2\).
Adding both equations: \((a - b) + (b - c) = 5k_1 + 5k_2\)
\(\Rightarrow a - c = 5(k_1 + k_2)\).
Since \(k_1 + k_2\) is an integer, \((a - c)\) is divisible by 5.
\(\Rightarrow (a, c) \in R\).
Therefore, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: A relation is an equivalence relation if it satisfies reflexivity, symmetry, and transitivity properties simultaneously.

Exam Tip: Always state the definition of divisibility clearly using an integer multiple (like \(5k\)) when checking for equivalence relations.

Q2. Let \(Q\) be the set of all rational numbers and \(R\) be the relation on \(Q\) defined by \(R = \{(x, y) : 1 + xy > 0\}\). Prove that \(R\) is reflexive and symmetric but not transitive.
Answer: 1. Reflexive:
For any \(x \in Q\), we have \(1 + x \cdot x = 1 + x^2\).
Since the square of any real or rational number is non-negative (\(x^2 \geq 0\)), we get \(1 + x^2 > 0\).
Therefore, \((x, x) \in R\) for all \(x \in Q\), so \(R\) is reflexive.
2. Symmetric:
Let \((x, y) \in R\) for \(x, y \in Q\).
Then \(1 + xy > 0\).
Since \(xy = yx\), this implies \(1 + yx > 0\).
Therefore, \((y, x) \in R\), so \(R\) is symmetric.
3. Transitive (Counterexample):
Let \(x = 10, y = -\frac{1}{2}, z = -\frac{1}{4}\) (all are rational numbers).
Check \((x, y)\): \(1 + (10)\left(-\frac{1}{2}\right) = 1 - 5 = -4\) (Wait, we need \(1 + xy > 0\), let us choose numbers carefully).
Let \(x = 0, y = 1, z = -10\):
Check \((0, 1)\): \(1 + (0)(1) = 1 > 0\) (\(\Rightarrow (0, 1) \in R\)).
Check \((1, -10)\): \(1 + (1)(-10) = -9 \not> 0\) (So \((1, -10) \notin R\), try another set of values).
Let \(x = -2, y = 0, z = -3\):
Check \((x, y) = (-2, 0)\): \(1 + (-2)(0) = 1 > 0\) (\(\in R\)).
Check \((y, z) = (0, -3)\): \(1 + (0)(-3) = 1 > 0\) (\(\in R\)).
Check \((x, z) = (-2, -3)\): \(1 + (-2)(-3) = 1 + 6 = 7 > 0\) (\(\in R\), this is valid, we need a failure).
Let us take \(x = -10, y = -\frac{1}{5}, z = 2\):
Check \((x, y) = \left(-10, -\frac{1}{5}\right)\): \(1 + (-10)\left(-\frac{1}{5}\right) = 1 + 2 = 3 > 0\) (\(\in R\)).
Check \((y, z) = \left(-\frac{1}{5}, 2\right)\): \(1 + \left(-\frac{1}{5}\right)(2) = 1 - \frac{2}{5} = \frac{3}{5} > 0\) (\(\in R\)).
Check \((x, z) = (-10, 2)\): \(1 + (-10)(2) = 1 - 20 = -19 \not> 0\) (\(\notin R\)).
Since \((-10, -\frac{1}{5}) \in R\) and \((-\frac{1}{5}, 2) \in R\) but \((-10, 2) \notin R\), the relation \(R\) is not transitive.
In simple words: To disprove transitivity, we only need to find a single counterexample where \((a,b)\) and \((b,c)\) belong to the relation, but \((a,c)\) does not.

Exam Tip: For disproving transitivity, carefully select numbers (often fractions or negative numbers) that violate the inequality condition for the third pair.

Q3. Let \(A = \{4, 5, 6\}\). Give an example of a relation on \(A\) which is:
(a) reflexive and symmetric but not transitive.
(b) reflexive and transitive but not symmetric.
(c) symmetric and transitive but not reflexive.
Answer:
(a) Reflexive and symmetric but not transitive:
\(R_1 = \{(4,4), (5,5), (6,6), (4,5), (5,4), (5,6), (6,5)\}\)
Check: Reflexive as all elements have identical pairs. Symmetric because inverses exist. Not transitive because \((4,5) \in R_1\) and \((5,6) \in R_1\), but \((4,6) \notin R_1\).
(b) Reflexive and transitive but not symmetric:
\(R_2 = \{(4,4), (5,5), (6,6), (4,5), (4,6)\}\)
Check: Reflexive and transitive. Not symmetric because \((4,5) \in R_2\) but \((5,4) \notin R_2\).
(c) Symmetric and transitive but not reflexive:
\(R_3 = \{(4,4), (4,5), (5,4), (5,5)\}\)
Check: Symmetric and transitive. Not reflexive because \((6,6) \notin R_3\).
In simple words: Construct relations by explicitly adding or omitting specific ordered pairs to test the boundaries of each property.

Exam Tip: When constructing examples on finite sets, always check all elements for reflexivity first, then verify symmetry and transitivity pairs.

Q4. Show that the relation \(R\) defined by \((a, b) R (c, d) \implies a + d = b + c\) on the set \(N \times N\) is an equivalence relation.
Answer: Let \(S = N \times N\).
1. Reflexive:
For any \((a, b) \in N \times N\), we have \(a + b = b + a\) (since addition is commutative in natural numbers).
Therefore, \((a, b) R (a, b)\), so \(R\) is reflexive.
2. Symmetric:
Let \((a, b) R (c, d)\).
Then \(a + d = b + c\).
This can be rewritten as \(c + b = d + a\), which means \(c + b = a + d\), or \((c, d) R (a, b)\).
Therefore, \(R\) is symmetric.
3. Transitive:
Let \((a, b) R (c, d)\) and \((c, d) R (e, f)\).
Then \(a + d = b + c\) and \(c + f = d + e\).
Adding both equations: \((a + d) + (c + f) = (b + c) + (d + e)\)
Cancelling \(c\) and \(d\) from both sides: \(a + f = b + e\).
This implies \((a, b) R (e, f)\).
Therefore, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Properties of equality and addition in natural numbers are used to prove reflexivity, symmetry, and transitivity for Cartesian product sets.

Exam Tip: For relations defined on Cartesian products like \(N \times N\), treat each element as an ordered pair and apply the given cross-sum condition.

Q5. Let \(f : A \to B\) be a function. Define a relation \(R\) in \(A\) given by \(R = \{(a, b) : f(a) = f(b)\}\). Is \(R\) an equivalence relation?
Answer: Yes, \(R\) is an equivalence relation.
1. Reflexive:
For any element \(a \in A\), we have \(f(a) = f(a)\).
Therefore, \((a, a) \in R\) for all \(a \in A\), so \(R\) is reflexive.
2. Symmetric:
Let \((a, b) \in R\).
Then \(f(a) = f(b)\).
This implies \(f(b) = f(a)\), which means \((b, a) \in R\).
Therefore, \(R\) is symmetric.
3. Transitive:
Let \((a, b) \in R\) and \((b, c) \in R\).
Then \(f(a) = f(b)\) and \(f(b) = f(c)\).
Combining these gives \(f(a) = f(c)\), which means \((a, c) \in R\).
Therefore, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Equality of function values naturally satisfies all three conditions required for an equivalence relation.

Exam Tip: Use the standard properties of equality (reflexive, symmetric, transitive nature of '=' symbol) to quickly establish equivalence for function-based relations.

Q6. Prove that the relation \(R\) in the set \(A = \{1, 2, 3, 4, 5\}\) given by \(R = \{(a, b) : |a - b| \text{ is even}\}\) is an equivalence relation.
Answer:
1. Reflexive:
For any \(a \in A\), \(|a - a| = 0\), which is an even number.
Therefore, \((a, a) \in R\) for all \(a \in A\), so \(R\) is reflexive.
2. Symmetric:
Let \((a, b) \in R\).
Then \(|a - b|\) is even.
Since \(|b - a| = |-(a - b)\} = |a - b|\), \(|b - a|\) is also even.
Therefore, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitive:
Let \((a, b) \in R\) and \((b, c) \in R\).
Then \(|a - b|\) is even and \(|b - c|\) is even.
This means \(a - b\) is even and \(b - c\) is even.
The sum of two even numbers is even: \((a - b) + (b - c) = a - c\) is even.
Therefore, \(|a - c|\) is even, which implies \((a, c) \in R\).
Thus, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: The absolute difference being even means the difference is a multiple of 2, which satisfies all equivalence conditions.

Exam Tip: Remember that the sum of two even integers is always even, which is crucial for proving transitivity with difference-based relations.

Q7. Show that the relation \(R\) in the set \(A\), \(A = \{x \in W, 0 \leq x \leq 12\}\) given by \(R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}\) is an equivalence relation. Also find the set of all elements related to 2.
Answer: Given \(A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\).
1. Reflexive:
For any \(a \in A\), \(|a - a| = 0\), which is a multiple of 4 (since \(0 = 4 \times 0\)).
Therefore, \((a, a) \in R\), so \(R\) is reflexive.
2. Symmetric:
Let \((a, b) \in R\).
Then \(|a - b|\) is a multiple of 4.
Since \(|b - a| = |a - b|\), \(|b - a|\) is also a multiple of 4.
Therefore, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitive:
Let \((a, b) \in R\) and \((b, c) \in R\).
Then \(|a - b|\) is a multiple of 4 and \(|b - c|\) is a multiple of 4.
Hence, \(a - b = 4k_1\) and \(b - c = 4k_2\) for some integers \(k_1, k_2\).
Adding them: \(a - c = 4(k_1 + k_2)\), which is a multiple of 4.
Therefore, \((a, c) \in R\), so \(R\) is transitive.
Thus, \(R\) is an equivalence relation.
Elements related to 2: Let \(x \in A\) be related to 2.
Then \(|x - 2|\) must be a multiple of 4, where \(x \in \{0, 1, 2, \dots, 12\}\).
Possible values of \(x\) such that \(|x - 2|\) is a multiple of 4:
- If \(x = 2\), \(|2 - 2| = 0\) (multiple of 4)
- If \(x = 6\), \(|6 - 2| = 4\) (multiple of 4)
- If \(x = 10\), \(|10 - 2| = 8\) (multiple of 4)
Therefore, the set of all elements related to 2 is \(\{2, 6, 10\}\).
In simple words: Elements related to a specific number are those whose difference with it is a multiple of the given number (4 in this case).

Exam Tip: For finite sets, finding the equivalence class is straightforward by testing every element of the set against the given condition.

Q8. Show that the relation \(S\) in the set \(A\), \(A = \{x \in Z : 0 \leq x \leq 12\}\) given by \(S = \{(a, b) : a, b \in A, |a - b| \text{ is divisible by } 4\}\) is an equivalence relation. Find the set of elements related to 1.
Answer: Given \(A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\).
1. Reflexive:
For any \(a \in A\), \(|a - a| = 0\), which is divisible by 4.
Therefore, \((a, a) \in S\), so \(S\) is reflexive.
2. Symmetric:
Let \((a, b) \in S\).
Then \(|a - b|\) is divisible by 4.
Since \(|b - a| = |a - b|\), \(|b - a|\) is also divisible by 4.
Therefore, \((b, a) \in S\), so \(S\) is symmetric.
3. Transitive:
Let \((a, b) \in S\) and \((b, c) \in S\).
Then \(|a - b|\) and \(|b - c|\) are both divisible by 4.
Thus, their difference/sum property shows \(|a - c|\) is divisible by 4.
Therefore, \((a, c) \in S\), so \(S\) is transitive.
Thus, \(S\) is an equivalence relation.
Elements related to 1: Let \(x \in A\) be related to 1.
Then \(|x - 1|\) must be divisible by 4, where \(x \in \{0, 1, \dots, 12\}\).
- If \(x = 1\), \(|1 - 1| = 0\) (divisible by 4)
- If \(x = 5\), \(|5 - 1| = 4\) (divisible by 4)
- If \(x = 9\), \(|9 - 1| = 8\) (divisible by 4)
Therefore, the set of all elements related to 1 is \(\{1, 5, 9\}\).
In simple words: The equivalence class of 1 contains all numbers in the set whose absolute difference with 1 is a multiple of 4.

Exam Tip: Double-check the boundaries of set \(A\) when listing equivalence class elements so you do not include numbers outside the defined domain.

Q9. Show that the relation \(R\) on the set \(I\) of all integers defined by \(R = \{(a, b) : a - b \text{ is divisible by } 3, a, b \in I\}\) is an equivalence relation.
Answer:
1. Reflexive:
For any integer \(a \in I\), \(a - a = 0\).
Since 0 is divisible by 3 (\(0 = 3 \times 0\)), \((a, a) \in R\) for all \(a \in I\).
Therefore, \(R\) is reflexive.
2. Symmetric:
Let \((a, b) \in R\) for \(a, b \in I\).
Then \(a - b\) is divisible by 3.
\(\Rightarrow a - b = 3k\) for some integer \(k\).
\(\Rightarrow b - a = 3(-k)\).
Since \(-k\) is an integer, \(b - a\) is divisible by 3.
\(\Rightarrow (b, a) \in R\).
Therefore, \(R\) is symmetric.
3. Transitive:
Let \((a, b) \in R\) and \((b, c) \in R\) for \(a, b, c \in I\).
Then \(a - b = 3k_1\) and \(b - c = 3k_2\) for integers \(k_1, k_2\).
Adding the two equations: \((a - b) + (b - c) = 3k_1 + 3k_2\)
\(\Rightarrow a - c = 3(k_1 + k_2)\).
Since \(k_1 + k_2\) is an integer, \(a - c\) is divisible by 3.
\(\Rightarrow (a, c) \in R\).
Therefore, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Divisibility by 3 follows the standard integer properties, making it an equivalence relation over the set of all integers.

Exam Tip: Write out the algebraic form \(3k\) explicitly for divisibility proofs to ensure full marks.

Q10. State the reason for the relation \(R\) in the set \(\{1, 2, 3\}\) given by \(R = \{(1, 2), (2, 1)\}\) not to be transitive.
Answer: The relation \(R = \{(1, 2), (2, 1)\}\) is not transitive because:\br /> We have \((1, 2) \in R\) and \((2, 1) \in R\).
For \(R\) to be transitive, \((1, 1)\) must belong to \(R\).
However, \((1, 1) \notin R\).
Therefore, the relation is not transitive.
In simple words: Transitivity requires that if \((a,b)\) and \((b,c)\) are present, then \((a,c)\) must also be present in the relation.

Exam Tip: Look for matching intermediate elements like 2 in \((1,2)\) and \((2,1)\) to check if the combined pair \((1,1)\) is present when testing transitivity.