CBSE Class 12 Mathematics Relations And Functions Class Test Set 05

Question 1. Show that the relation \(R\) in the set \(\mathbb{Z}\) of integers given by \(R = \{(a, b) : 2 \text{ divides } (a - b)\}\) for all \(a, b \in \mathbb{Z}\) is an equivalence relation.
Answer: To show that \(R\) is an equivalence relation, we must prove that it is reflexive, symmetric, and transitive.
1. Reflexive:
For any integer \(a \in \mathbb{Z}\), we have \(a - a = 0\).
Since \(0\) is divisible by \(2\) (\(0 = 2 \times 0\)), we have \((a, a) \in R\) for all \(a \in \mathbb{Z}\).
Thus, \(R\) is reflexive.
2. Symmetric:
Let \((a, b) \in R\) for some integers \(a, b \in \mathbb{Z}\).
This means \(2\) divides \((a - b)\), so \(a - b = 2k\) for some integer \(k\).
Then, \(b - a = -(a - b) = -2k = 2(-k)\), which shows that \(2\) divides \((b - a)\).
Thus, \((b, a) \in R\).
Hence, \(R\) is symmetric.
3. Transitive:
Let \((a, b) \in R\) and \((b, c) \in R\) for integers \(a, b, c \in \mathbb{Z}\).
This means \(2\) divides \((a - b)\) and \(2\) divides \((b - c)\).
So, \(a - b = 2k_1\) and \(b - c = 2k_2\) for some integers \(k_1, k_2\).
Adding these two equations, we get:
\((a - b) + (b - c) = 2k_1 + 2k_2\)
\(a - c = 2(k_1 + k_2)\)
This shows that \(2\) divides \((a - c)\), so \((a, c) \in R\).
Hence, \(R\) 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 definition of divisibility by 2.

Exam Tip: Always clearly state and check all three conditions (reflexive, symmetric, transitive) separately with clear algebraic definitions.

Question 2. \(L\) is the set of all lines in a plane and \(R\) is the relation in \(L\) given by \(R = \{(l_1, l_2) : l_1 \perp l_2, \forall l_1, l_2 \in L\}\). Is \(R\) an equivalence relation?
Answer: Let us check the three properties for relation \(R\):
1. Reflexive:
A line cannot be perpendicular to itself because no line is perpendicular to itself (\(l_1 \not\perp l_1\)).
Thus, \((l_1, l_1) \notin R\).
Therefore, \(R\) is not reflexive.
2. Symmetric:
Let \((l_1, l_2) \in R\). This means line \(l_1\) is perpendicular to line \(l_2\) (\(l_1 \perp l_2\)).
If \(l_1 \perp l_2\), then clearly line \(l_2\) is also perpendicular to line \(l_1\) (\(l_2 \perp l_1\)).
Thus, \((l_2, l_1) \in R\).
Therefore, \(R\) is symmetric.
3. Transitive:
Let \((l_1, l_2) \in R\) and \((l_2, l_3) \in R\).
This means \(l_1 \perp l_2\) and \(l_2 \perp l_3\).
If \(l_1\) is perpendicular to \(l_2\) and \(l_2\) is perpendicular to \(l_3\), then \(l_1\) and \(l_3\) are parallel to each other, not perpendicular (\(l_1 \parallel l_3\)).
Thus, \((l_1, l_3) \notin R\).
Therefore, \(R\) is not transitive.
Since \(R\) is not reflexive and not transitive, it is not an equivalence relation.
In simple words: Perpendicularity is symmetric, but a line is never perpendicular to itself, and two lines perpendicular to the same line are parallel, making it neither reflexive nor transitive.

Exam Tip: A relation must satisfy all three conditions (reflexive, symmetric, and transitive) to be an equivalence relation. Failure of even one condition means it is not.

Question 3. State the reason for the relation \(R\) in the set \(\{1, 2, 3\}\) given by \(\{(1, 2), (2, 1)\}\) not to be transitive.
Answer: For a relation to be transitive, whenever \((a, b) \in R\) and \((b, c) \in R\), we must have \((a, c) \in R\).
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, \((1, 1) \notin R\).
Therefore, the relation \(R\) is not transitive.
In simple words: Even though we have pairs connecting 1 to 2 and 2 to 1, the required direct connection from 1 to 1 (\((1, 1)\)) is missing.

Exam Tip: When checking transitivity, always look for chained pairs like \((a, b)\) and \((b, c)\) and verify whether \((a, c)\) is present in the set.

Question 4. Let \(A = \{1, 2, 3, 4, 5\}\) and \(R = \{(a, b) : b = a + 1, \forall a, b \in A\}\). Check whether the above relation is reflexive, symmetric or transitive.
Answer: First, let us write the relation in roster form based on the condition \(b = a + 1\):
\(R = \{(1, 2), (2, 3), (3, 4), (4, 5)\}\)
Now, let us test for the properties:
1. Reflexive:
For a relation to be reflexive, \((a, a) \in R\) for all \(a \in A\).
Here, \((1, 1), (2, 2), (3, 3), (4, 4), (5, 5)\) are not in \(R\).
For example, \(1 = 1 + 1\) is false, so \((1, 1) \notin R\).
Thus, \(R\) is not reflexive.
2. Symmetric:
For a relation to be symmetric, whenever \((a, b) \in R\), then \((b, a) \in R\).
Here, \((1, 2) \in R\), but \((2, 1) \notin R\).
Thus, \(R\) is not symmetric.
3. Transitive:
For a relation to be transitive, whenever \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\).
Here, \((1, 2) \in R\) and \((2, 3) \in R\), but the pair \((1, 3) \notin R\) (since \(3 \neq 1 + 1\)).
Thus, \(R\) is not transitive.
Therefore, the relation \(R\) is neither reflexive, nor symmetric, nor transitive.
In simple words: By listing out the valid pairs where the second number is one more than the first, we can easily test and show that none of the three relation properties hold true.

Exam Tip: Converting a set-builder relation into roster form makes checking reflexive, symmetric, and transitive properties much easier and less prone to errors.