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Q8:
Show that the relation $R$ in the set $A = \{1, 2, 3, 4, 5\}$ given by $R = \{(a, b) : |a – b|$ is even\}, is an equivalence relation. Show that all the elements of $\{1, 3, 5\}$ are related to each other and all the elements of $\{2, 4\}$ are related to each other. But no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

Solution :

Step 1: Establishing the Relation Definition

Given set $A = \{1, 2, 3, 4, 5\}$. The relation $R$ is defined as:

$R = \{(a, b) : a, b \in A, |a - b| \text{ is even}\}$

Step 2: Proving the Equivalence Relation

To prove $R$ is an equivalence relation, we must demonstrate that it is reflexive, symmetric, and transitive.

(i) Reflexivity: For any $a \in A$, we observe $|a - a| = 0$. Since $0$ is an even number, $(a, a) \in R$ for all $a \in A$. [Per the definition of Reflexive Relations].

(ii) Symmetry: Let $(a, b) \in R$. This implies $|a - b|$ is even. Since $|a - b| = |-(b - a)| = |b - a|$, then $|b - a|$ is also even. Thus, $(b, a) \in R$. [Per the definition of Symmetric Relations].

(iii) Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. Then $|a - b| = 2k_1$ and $|b - c| = 2k_2$ for some integers $k_1, k_2$. We express $a - c = (a - b) + (b - c)$. Since the sum of two even numbers is even, $(a - c)$ is even, hence $|a - c|$ is even. Thus, $(a, c) \in R$. [Per the definition of Transitive Relations].

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Step 3: Analyzing Subsets and Equivalence Classes

We analyze the two subsets $S_1 = \{1, 3, 5\}$ and $S_2 = \{2, 4\}$.

SubsetLogic
$S_1 = \{1, 3, 5\}$For any $a, b \in \{1, 3, 5\}$, the difference $a - b$ is always even (e.g., $|3-1|=2, |5-3|=2, |5-1|=4$). All are related.
$S_2 = \{2, 4\}$For any $a, b \in \{2, 4\}$, the difference $|4-2|=2$, which is even. All are related.

To show that no element of $S_1$ is related to $S_2$, let $a \in \{1, 3, 5\}$ (odd) and $b \in \{2, 4\}$ (even). The difference $a - b$ is always odd (e.g., $1-2 = -1, 3-4 = -1, 5-2 = 3$). Since an odd number is not even, $|a - b|$ is not even. Therefore, $(a, b) \notin R$.

Step 4: Visualizing the Equivalence Classes

1, 3, 5 2, 4 Equivalence Class [1] Equivalence Class [2] No Relation

Final Solution: The relation is proven to be an equivalence relation as it satisfies reflexivity, symmetry, and transitivity. The set $A$ is partitioned into two disjoint equivalence classes: $\{1, 3, 5\}$ and $\{2, 4\}$, where cross-set relations are impossible due to the parity properties of integers.


More Questions from Class 12 Mathematics Relations and Functions EXERCISE 1.1


CBSE Solutions for Class 12 Mathematics Relations and Functions


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