CBSE Class 12 Mathematics Relations And Functions Assignment Set 10

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. Reflexivity: For any integer \(a \in Z\), we have \(a - a = 0\). Since 0 is divisible by 5, \((a, a) \in R\) for all \(a \in Z\). Thus, \(R\) is reflexive.
2. Symmetry: Let \((a, b) \in R\). This means \(a - b\) is divisible by 5, so \(a - b = 5k\) for some integer \(k\). Then \(b - a = -(a - b) = 5(-k)\), which means \(b - a\) is also divisible by 5. Thus, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(a - b\) is divisible by 5 and \(b - c\) is divisible by 5. Let \(a - b = 5k_1\) and \(b - c = 5k_2\) for integers \(k_1, k_2\). Adding both equations gives \((a - b) + (b - c) = 5k_1 + 5k_2\), which simplifies to \(a - c = 5(k_1 + k_2)\). This shows that \(a - c\) is divisible by 5, so \((a, c) \in R\). Thus, \(R\) 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, all of which hold here based on divisibility by 5.

Exam Tip: Clearly state the definition of reflexivity, symmetry, and transitivity with algebraic expressions using variables like \(k, k_1, k_2\).

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. Reflexivity: For any \(x \in Q\), \(1 + x \cdot x = 1 + x^2\). Since \(x^2 \ge 0\), we have \(1 + x^2 > 0\). Thus, \((x, x) \in R\) for all \(x \in Q\), so \(R\) is reflexive.
2. Symmetry: Let \((x, y) \in R\). Then \(1 + xy > 0\). Since \(xy = yx\), we have \(1 + yx > 0\), which implies \((y, x) \in R\). Thus, \(R\) is symmetric.
3. Transitivity: To show \(R\) is not transitive, let us find a counterexample. Let \(x = 2, y = -\frac{1}{2}, z = -4\).
Check \((x, y)\): \(1 + (2)\left(-\frac{1}{2}\right) = 1 - 1 = 0\). Wait, let us choose values such that \(1 + xy > 0\) and \(1 + yz > 0\), but \(1 + xz \le 0\).
Let \(x = 10, y = \frac{1}{2}, z = -\frac{1}{3}\).
\(1 + xy = 1 + (10)\left(\frac{1}{2}\right) = 6 > 0\) (so \((x, y) \in R\)).
\(1 + yz = 1 + \left(\frac{1}{2}\right)\left(-\frac{1}{3}\right) = 1 - \frac{1}{6} = \frac{5}{6} > 0\) (so \((y, z) \in R\)).
Now check \((x, z)\): \(1 + xz = 1 + (10)\left(-\frac{1}{3}\right) = 1 - \frac{10}{3} = -\frac{7}{3} \le 0\) (so \((x, z) \notin R\)).
Thus, \(R\) is not transitive.
In simple words: Reflexivity and symmetry are proved directly using properties of numbers, while transitivity is disproved by finding a counterexample.

Exam Tip: For disproving transitivity, always explicitly construct a suitable set of counterexample numbers satisfying the first two relations.

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) \(R_1 = \{(4, 4), (5, 5), (6, 6), (4, 5), (5, 4), (5, 6), (6, 5)\}\)
Here, \((4, 5) \in R_1\) and \((5, 6) \in R_1\), but \((4, 6) \notin R_1\). Thus, it is not transitive, while clearly reflexive and symmetric.
(b) \(R_2 = \{(4, 4), (5, 5), (6, 6), (4, 5), (4, 6)\}\)
Here, \((4, 5) \in R_2\), but \((5, 4) \notin R_2\), so it is not symmetric. It is reflexive and transitive.
(c) \(R_3 = \{(4, 4), (4, 5), (5, 4), (5, 5)\}\)
Here, element \(6\) has no relation like \((6, 6)\), so it is not reflexive. It is symmetric and transitive.
In simple words: Construct relations by selectively including pairs to satisfy specific properties while violating the remaining one.

Exam Tip: Always verify all elements of the set \(A\) when checking reflexivity.

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: 1. Reflexivity: For any \((a, b) \in N \times N\), we have \(a + b = b + a\). Therefore, \((a, b) R (a, b)\). Thus, \(R\) is reflexive.
2. Symmetry: Let \((a, b) R (c, d)\). Then \(a + d = b + c\). This can be rewritten as \(c + b = d + a\), which means \(c + b = a + d\), so \((c, d) R (a, b)\). Thus, \(R\) is symmetric.
3. Transitivity: Let \((a, b) R (c, d)\) and \((c, d) R (e, f)\). Then:
\(a + d = b + c\) (Equation 1)
\(c + f = d + e\) (Equation 2)
Adding Equation 1 and Equation 2:
\((a + d) + (c + f) = (b + c) + (d + e)\)
\(a + f + c + d = b + e + c + d\)
Cancelling \(c + d\) from both sides:
\(a + f = b + e\)
This implies \((a, b) R (e, f)\). Thus, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Treat ordered pairs like equations of sums and apply standard algebraic addition and cancellation.

Exam Tip: Use standard cross-addition of equations to easily prove transitivity on cartesian product sets.

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. Reflexivity: For any \(a \in A\), \(f(a) = f(a)\) is always true. Thus, \((a, a) \in R\), so \(R\) is reflexive.
2. Symmetry: Let \((a, b) \in R\). Then \(f(a) = f(b)\), which implies \(f(b) = f(a)\). Thus, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(f(a) = f(b)\) and \(f(b) = f(c)\). Therefore, \(f(a) = f(c)\), 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: Since function equality satisfies reflexivity, symmetry, and transitivity of real numbers, the induced relation is also an equivalence relation.

Exam Tip: This is a standard theoretical proof; rely directly on the properties of equality of function outputs.

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. Reflexivity: For any \(a \in A\), \(|a - a| = 0\), which is an even number. Thus, \((a, a) \in R\) for all \(a \in A\), so \(R\) is reflexive.
2. Symmetry: Let \((a, b) \in R\). Then \(|a - b|\) is even. Since |a - b| = |b - a|, \(|b - a|\) is also even. Thus, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(|a - b|\) is even and \(|b - c|\) is even. The sum or difference of two even integers is even. Therefore, \((a - b) + (b - c) = a - c\) is even, which means \(|a - c|\) is even. Thus, \((a, c) \in R\), so \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: The difference between two numbers being even maintains reflexivity, symmetry, and transitivity.

Exam Tip: Mention clearly that the sum of two even integers is always even when proving transitivity.

Q7. Show that the relation \(R\) in the set \(A\), \(A = \{x \in W, 0 \le x \le 12\}\) 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: The set is \(A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\).
1. Reflexivity: For any \(a \in A\), \(|a - a| = 0\), which is a multiple of 4. Thus, \((a, a) \in R\), so \(R\) is reflexive.
2. Symmetry: Let \((a, b) \in R\). Then \(|a - b|\) is a multiple of 4. Since \(|a - b| = |b - a|\), \(|b - a|\) is also a multiple of 4. Thus, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(|a - b| = 4k_1\) and \(|b - c| = 4k_2\). Since \(a - c = (a - b) + (b - c)\), it is a multiple of 4. Thus, \((a, c) \in R\), so \(R\) is transitive.
Therefore, \(R\) is an equivalence relation.
Finding elements related to 2: Let \(x\) be related to 2, so \(|x - 2|\) must be a multiple of 4, where \(x \in A\).
Possible values for \(x\):
If \(|x - 2| = 0 \implies x = 2\)
If \(|x - 2| = 4 \implies x - 2 = 4 \implies x = 6\) or \(2 - x = 4 \implies x = -2\) (not in \(A\))
If \(|x - 2| = 8 \implies x - 2 = 8 \implies x = 10\)
Thus, the set of all elements related to 2 is \(\{2, 6, 10\}\).
In simple words: Elements related to 2 are those numbers in the set whose difference from 2 is divisible by 4.

Exam Tip: Be careful to restrict the final elements strictly within the given domain set \(A\).

Q8. Show that the relation \(S\) in the set \(A\), \(A = \{x \in Z : 0 \le x \le 12\}\) 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: The set is \(A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\).
1. Reflexivity: For any \(a \in A\), \(|a - a| = 0\), which is divisible by 4. Thus, \((a, a) \in S\), so \(S\) is reflexive.
2. Symmetry: Let \((a, b) \in S\). Then \(|a - b|\) is divisible by 4, which means \(|b - a|\) is divisible by 4. Thus, \((b, a) \in S\), so \(S\) is symmetric.
3. Transitivity: Let \((a, b) \in S\) and \((b, c) \in S\). Then \(|a - b|\) and \(|b - c|\) are divisible by 4. Hence, their sum/difference is divisible by 4, making \(|a - c|\) divisible by 4. Thus, \((a, c) \in S\), so \(S\) is transitive.
Therefore, \(S\) is an equivalence relation.
Finding elements related to 1: Let \(x \in A\) such that \(|x - 1|\) is divisible by 4.
\(|x - 1| = 0 \implies x = 1\)
\(|x - 1| = 4 \implies x - 1 = 4 \implies x = 5\) or \(1 - x = 4 \implies x = -3\) (not in \(A\))
\(|x - 1| = 8 \implies x - 1 = 8 \implies x = 9\)
Thus, the set of elements related to 1 is \(\{1, 5, 9\}\).
In simple words: The equivalence class of 1 contains all numbers in range whose difference with 1 is a multiple of 4.

Exam Tip: Clearly check the boundary conditions of the set \(A\) when listing equivalence classes.

Q9. Show that the relation \(R\) on the set \(Z\) of all integers defined by \(R = \{(a, b) : (a - b) \text{ is divisible by } 3, a, b \in Z\}\) is an equivalence relation.
Answer: 1. Reflexivity: For any integer \(a \in Z\), \(a - a = 0\), which is divisible by 3. Thus, \((a, a) \in R\), so \(R\) is reflexive.
2. Symmetry: Let \((a, b) \in R\). Then \(a - b = 3k\) for some integer \(k\). Then \(b - a = -3k = 3(-k)\), which means \(b - a\) is divisible by 3. Thus, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(a - b = 3k_1\) and \(b - c = 3k_2\) for integers \(k_1, k_2\). Adding both gives \(a - c = 3(k_1 + k_2)\), which is divisible by 3. Thus, \((a, c) \in R\), so \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Divisibility by 3 behaves identically to divisibility by 5 or 4 in terms of establishing equivalence relations.

Exam Tip: Follow the standard three-step format meticulously for all divisibility-based equivalence relation questions.

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: A relation is transitive if whenever \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\).
Here, \((1, 2) \in R\) and \((2, 1) \in R\). For \(R\) to be transitive, we must have \((1, 1) \in R\). However, \((1, 1) \notin R\). Therefore, the relation is not transitive.
In simple words: A chain of pairs like (1, 2) and (2, 1) requires the direct pair (1, 1) to be present for transitivity, which is missing here.

Exam Tip: Explicitly state the missing ordered pair that causes the failure of transitivity.