CBSE Class 12 Mathematics Relations And Functions Assignment Set 11

XII - Mathematics Assignment - Relations And Functions

Question 1. 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 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. Symmetric: Let \((a, b) \in R\) for some integers \(a, b \in Z\). This means \((a - b)\) is divisible by 5, so \(a - b = 5k\) for some integer \(k\). Then \(b - a = -5k = 5(-k)\), which means \((b - a)\) is also divisible by 5. Therefore, \((b, a) \in R\). Thus, \(R\) is symmetric.
3. Transitive: Let \((a, b) \in R\) and \((b, c) \in R\) for integers \(a, b, c \in Z\). 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 implies \(a - c = 5(k_1 + k_2)\). Thus, \((a - c)\) 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, which are all proven here using the property of divisibility by 5.

Exam Tip: Always explicitly write out the definitions for reflexivity, symmetry, and transitivity with general elements before substituting the given condition.

Question 2. 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 need \(1 + xx > 0\), which means \(1 + x^2 > 0\). Since the square of any real rational number is non-negative (\(x^2 \ge 0\)), \(1 + x^2 \ge 1 > 0\) is always true. Thus, \((x, x) \in R\), so \(R\) is reflexive.
2. Symmetric: Let \((x, y) \in R\). Then \(1 + xy > 0\). Since multiplication of rational numbers is commutative, \(xy = yx\), which implies \(1 + yx > 0\). Thus, \((y, x) \in R\), so \(R\) is symmetric.
3. Transitive (Counterexample): To show it is not transitive, let us choose rational numbers \(x = 10\), \(y = \frac{1}{2}\), and \(z = -\frac{1}{2}\).
Check \((10, \frac{1}{2})\): \(1 + (10)(\frac{1}{2}) = 1 + 5 = 6 > 0\) (True, so \((10, \frac{1}{2}) \in R\)).
Check \((\frac{1}{2}, -\frac{1}{2})\): \(1 + (\frac{1}{2})(-\frac{1}{2}) = 1 - \frac{1}{4} = \frac{3}{4} > 0\) (True, so \((\frac{1}{2}, -\frac{1}{2}) \in R\)).
Check \((10, -\frac{1}{2})\): \(1 + (10)(-\frac{1}{2}) = 1 - 5 = -4 \ngtr 0\) (False, so \((10, -\frac{1}{2}) \notin R\)).
Therefore, \(R\) is not transitive.
In simple words: Reflexivity and symmetry hold because squares are positive and multiplication is commutative, but transitivity fails when tested with specific numbers.

Exam Tip: For disproving transitivity, always use carefully chosen counterexamples with fractions or negative numbers.

Question 3. 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: Let \(R_1 = \{(4, 4), (5, 5), (6, 6), (4, 5), (5, 4), (5, 6), (6, 5)\}\).
It is reflexive (contains all identity pairs) and symmetric, but 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: Let \(R_2 = \{(4, 4), (5, 5), (6, 6), (4, 5)\}\).
It is reflexive and transitive, but not symmetric because \((4, 5) \in R_2\) but \((5, 4) \notin R_2$.
(c) Symmetric and transitive but not reflexive: Let \(R_3 = \{(4, 4), (4, 5), (5, 4), (5, 5)\}\).
It is symmetric and transitive, but not reflexive because \((6, 6) \notin R_3$.
In simple words: Construct relations by explicitly picking pairs from set \(A\) to satisfy the required properties while violating the forbidden one.

Exam Tip: Always check all elements of the set for reflexivity and all pairs for symmetry and transitivity when constructing custom relations.

Question 4. 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 \((a, b), (c, d), (e, f) \in N \times N$.
1. Reflexive: For any \((a, b) \in N \times N\), we have \(a + b = b + a\) by the commutative property of addition in natural numbers. Thus, \((a, b) R (a, b)\). Hence, \(R\) 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 equivalently \((c, d) R (a, b)\). Hence, \(R\) is symmetric.
3. Transitive: Let \((a, b) R (c, d)\) and \((c, d) R (e, f)\). Then:\br /> \(a + d = b + c \implies a + d - b - c = 0\) (Equation 1)
\(c + f = d + e \implies c + f - d - e = 0\) (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 \((a, b) R (e, f)\). Hence, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: By using basic algebraic properties of addition, we show that the cross-sum condition behaves consistently for reflexivity, symmetry, and transitivity.

Exam Tip: When dealing with Cartesian products like \(N \times N\), treat each element as an ordered pair \((a, b)\) and apply the given definition directly.

Question 5. 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 \(a \in A\), we have \(f(a) = f(a)\). Therefore, \((a, a) \in R\). Thus, \(R\) is reflexive.
2. Symmetric: Let \((a, b) \in R\). Then \(f(a) = f(b)\), which implies \(f(b) = f(a)\). Therefore, \((b, a) \in R\). Thus, \(R\) 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 \(f(a) = f(c)\). Therefore, \((a, c) \in R\). Thus, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Equality of function values is inherently reflexive, symmetric, and transitive, making the resulting relation an equivalence relation.

Exam Tip: This is a standard theoretical proof. Remember that equality \((=)\) is always an equivalence relation, which makes this proof straightforward.

Question 6. 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: For any \(a, b, c \in A\):
1. Reflexive: For any \(a \in A\), \(|a - a| = 0\), which is an even number. Thus, \((a, a) \in R\), so \(R\) is reflexive.
2. Symmetric: 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. Transitive: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(|a - b|\) is even and \(|b - c|\) is even. This means \((a - b)\) and \((b - c)\) are either both even or both odd. The sum of two numbers with the same parity is always even, so \((a - b) + (b - c) = a - c\) is even. Thus, \(|a - c|\) is even, which means \((a, c) \in R\). Hence, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: The absolute difference being even means the numbers have the same parity (both odd or both even), satisfying all three equivalence conditions.

Exam Tip: Use the property that the sum of two integers of the same parity is always even to prove transitivity cleanly.

Question 7. 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 \(A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\).
1. Reflexive: For any \(a \in A\), \(|a - a| = 0\), which is a multiple of 4 (since \(0 = 4 \times 0\)). Thus, \(\(a, a) \in R\).
2. Symmetric: 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, \((b, a) \in R\).
3. Transitive: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(|a - b| = 4k_1\) and \(|b - c| = 4k_2\) for some integers \(k_1, k_2\). This implies \(a - b = \pm 4k_1\) and \(b - c = \pm 4k_2\). Adding them gives \(a - c = \pm 4(k_1 \pm k_2)\), so \(|a - c|\) is a multiple of 4. Thus, \((a, c) \in R\).
Therefore, \(R\) is an equivalence relation.
To find all elements related to 2 (let this be set \([2]\)):
We need \(|a - 2|\) to be a multiple of 4 for \(a \in A\):
- If \(a = 2\), \(|2 - 2| = 0\) (multiple of 4)
- If \(a = 6\), \(|6 - 2| = 4\) (multiple of 4)
- If \(a = 10\), \(|10 - 2| = 8\) (multiple of 4)
Thus, the set of elements related to 2 is \(\{2, 6, 10\}\).
In simple words: Multiples of 4 difference means elements differ by 0, 4, 8, etc., allowing us to easily list elements related to 2 within the given range.

Exam Tip: For equivalence classes, test every element in the given finite set against the relation condition to ensure completeness.

Question 8. Show that the relation \(S\) in the set \(A = \{x \in Z : 0 \le x \le 12\}\) 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 \(A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\).
1. Reflexive: For any \(a \in A\), \(|a - a| = 0\), which is divisible by 4. Thus, \((a, a) \in S\).
2. Symmetric: Let \((a, b) \in S\). Then \(|a - b|\) is divisible by 4. Since \(|a - b| = |b - a|\), \(|b - a|\) is also divisible by 4. Thus, \((b, a) \in S\).
3. Transitive: Let \((a, b) \in S\) and \((b, c) \in S\). Then \(|a - b|\) and \(|b - c|\) are divisible by 4, meaning their sum \(|a - c|\) is also divisible by 4. Thus, \((a, c) \in S\).
Hence, \(S\) is an equivalence relation.
To find the set of elements related to 1 (let this be set \([1]\)):
We need \(|a - 1|\) to be divisible by 4 for \(a \in A\):
- If \(a = 1\), \(|1 - 1| = 0\) (divisible by 4)
- If \(a = 5\), \(|5 - 1| = 4\) (divisible by 4)
- If \(a = 9\), \(|9 - 1| = 8\) (divisible by 4)
Thus, the set of elements related to 1 is \(\{1, 5, 9\}\).
In simple words: The equivalence class for 1 consists of integers between 0 and 12 whose difference from 1 is a multiple of 4.

Exam Tip: Clearly write out the set \(A\) first so you do not miss any boundary elements when finding equivalence classes.

Question 9. 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: For any integers \(a, b, c \in I\):
1. Reflexive: For any \(a \in I\), \(a - a = 0\), which is divisible by 3. Thus, \((a, a) \in R\), so \(R\) is reflexive.
2. Symmetric: Let \((a, b) \in R\). Then \(a - b\) is divisible by 3, so \(a - b = 3k\) for some integer \(k\). Then \(b - a = -3k = 3(-k)\), which means \(b - a\) is also divisible by 3. Thus, \((b, a) \in R\), so \(R\) is symmetric.
3. Transitive: Let \((a, b) \in R\) and \((b, c) \in R\). Then \(a - b = 3k_1\) and \(b - c = 3k_2\) for some integers \(k_1, k_2\). Adding both equations gives \((a - b) + (b - c) = 3k_1 + 3k_2\), which simplifies to \(a - c = 3(k_1 + k_2)\). Thus, \(a - c\) is divisible by 3, so \((a, c) \in R\). Hence, \(R\) 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, satisfying all three conditions for an equivalence relation.

Exam Tip: Follow the standard template for divisibility relations: express the condition as a multiple of the given number \((\text{e.g., } 3k)\) and combine equations for transitivity.

Question 10. 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 \((a, c) \in R\).
In the given relation \(R = \{(1, 2), (2, 1)\}\):
We have \((1, 2) \in R\) and \((2, 1) \in R\).
For \(R\) to be transitive, we must have \((1, 1) \in R\). However, \((1, 1) \notin R\).
Similarly, we have \((2, 1) \in R\) and \((1, 2) \in R\), but \((2, 2) \notin R\).
Therefore, \(R\) is not transitive.
In simple words: Transitivity fails because the intermediate links exist (\((1, 2)\) and \((2, 1)\)), but the required direct connection \((1, 1)\) is missing.

Exam Tip: When checking for transitivity violations, look for pairs ending and starting with the same element (like \((a, b)\) and \((b, c)\)) and check if \((a, c)\) is present.