CBSE Class 12 Mathematics Relations And Functions Class Test Set 04

Q1. Show that the relation \(R\) in the set \(\mathbb{Z}\) of integers given by \(R = \{(a, b) : 2 \text{ divides } (a - b), \forall a, b \in \mathbb{Z}\}\) is an equivalence relation.
Answer: To show that \(R\) is an equivalence relation, we must prove that it is reflexive, symmetric, and transitive.
1. Reflexivity:
For any integer \(a \in \mathbb{Z}\), we have \(a - a = 0\).
Since 0 is divisible by 2 (\(0 = 2 \times 0\)), \((a, a) \in R\) for all \(a \in \mathbb{Z}\).
Therefore, \(R\) is reflexive.
2. Symmetry:
Let \((a, b) \in R\) for some integers \(a, b \in \mathbb{Z}\).
This means 2 divides \((a - b)\), which implies \(a - b = 2k\) for some integer \(k\).
Then, \(b - a = -(a - b) = -2k = 2(-k)\).
Since \(-k\) is an integer, 2 divides \((b - a)\), so \((b, a) \in R\).
Therefore, \(R\) is symmetric.
3. Transitivity:
Let \((a, b) \in R\) and \((b, c) \in R\) for some integers \(a, b, c \in \mathbb{Z}\).
This means \(a - b = 2k_1\) and \(b - c = 2k_2\) for some integers \(k_1\) and \(k_2\).
Adding the two equations, we get \((a - b) + (b - c) = 2k_1 + 2k_2\), which simplifies to \(a - c = 2(k_1 + k_2)\).
Since \(k_1 + k_2\) is an integer, 2 divides \((a - c)\), so \((a, c) \in R\).
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. Here, divisibility by 2 satisfies all three conditions.

Exam Tip: Clearly state the definition of reflexivity, symmetry, and transitivity with general elements before proving them for the given relation.

Q2. \(L\) is the set of all lines in a plane and \(R\) is the relation in \(L\) defined 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. Reflexivity:
A line cannot be perpendicular to itself because no line makes a 90 degree angle with itself (\(l_1 \not\perp l_1\)).
Therefore, \((l_1, l_1) \notin R\). Thus, \(R\) is not reflexive.
2. Symmetry:
Let \((l_1, l_2) \in R\). This means line \(l_1\) is perpendicular to line \(l_2\) (\(l_1 \perp l_2\)).
If \(l_1\) is perpendicular to \(l_2\), then obviously \(l_2\) is also perpendicular to \(l_1\) (\(l_2 \perp l_1\)).
Therefore, \((l_2, l_1) \in R\). Thus, \(R\) is symmetric.
3. Transitivity:
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 line \(l_1\) is perpendicular to line \(l_2\) and \(l_2\) is perpendicular to line \(l_3\), then \(l_1\) and \(l_3\) are parallel to each other, not perpendicular (\(l_1 \parallel l_3\)).
Therefore, \((l_1, l_3) \notin R\). Thus, \(R\) is not transitive.
Since \(R\) is not reflexive and not transitive, it is not an equivalence relation.
In simple words: Perpendicularity is symmetric (if line A is perpendicular to B, B is perpendicular to A), but a line is never perpendicular to itself, and two lines perpendicular to the same line are parallel, not perpendicular.

Exam Tip: Provide counter-examples or geometric reasoning when showing a relation fails to be reflexive, symmetric, or transitive.

Q3. State the reason for the relation \(R\) in the set \(\{1, 2, 3\}\) given by \(\{(1, 2), (2, 1)\}\) not to be transitive.
Answer: A relation \(R\) on a set \(A\) is transitive if whenever \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\) for all \(a, b, c \in A\).
Here, we have \((1, 2) \in R\) and \(\text{(2, 1)} \in R\).
For \(R\) to be transitive, we must have \((1, 1) \in R\).
However, \((1, 1) \notin R\).
Similarly, \((2, 2) \notin R\), which also violates transitivity considering the pairs in reverse order.
Therefore, the relation is not transitive.
In simple words: A relation fails to be transitive when a connection links a to b, and b to c, but there is no direct link from a to c.

Exam Tip: Always look for matching intermediate elements like \((a, b)\) and \((b, c)\) to test for the required pair \((a, c)\).

Q4. 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 list the elements of the relation \(R\) by substituting values from set \(A = \{1, 2, 3, 4, 5\}\) into \(b = a + 1\):
\(R = \{(1, 2), (2, 3), (3, 4), (4, 5)\}\)
1. Reflexivity:
For \(R\) to be reflexive, \((a, a)\) must belong to \(R\) for all \(a \in A\).
Here, \((1, 1), (2, 2), (3, 3), (4, 4), (5, 5)\) do not belong to \(R\).
Therefore, \(R\) is not reflexive.
2. Symmetry:
For \(R\) to be symmetric, if \((a, b) \in R\), then \((b, a)\) must belong to \(R\).
We have \((1, 2) \in R\), but \((2, 1) \notin R\).
Therefore, \(R\) is not symmetric.
3. Transitivity:
For \(R\) to be transitive, if \((a, b) \in R\) and \((b, c) \in R\), then \((a, c)\) must belong to \(R\).
We have \((1, 2) \in R\) and \((2, 3) \in R\), but \((1, 3) \notin R\).
Therefore, \(R\) is not transitive.
In simple words: Since no element is related to itself, no reverse pairs exist, and chained pairs do not connect directly, the relation is neither reflexive, symmetric, nor transitive.

Exam Tip: Converting a rule-based relation into roster form (listing the pairs) makes checking properties much easier and less prone to errors.