Question

Show 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.

Hint:

For relations defined using \(\mathbf{|a-b|}\), check parity (odd/even). Numbers with the {same parity (both even or both odd)} have an even difference. Thus, the equivalence classes here are: \[ \{1,3,5\}, \quad \{2,4\}. \]

Answer 1

Concept: A relation \(R\) on a set \(A\) is called an equivalence relation if it satisfies the following three properties:

• Reflexive: \( (a,a) \in R \) for all \(a \in A\)
• Symmetric: If \( (a,b) \in R \), then \( (b,a) \in R \)
• Transitive: If \( (a,b) \in R \) and \( (b,c) \in R \), then \( (a,c) \in R \)

Here, the relation is defined by \( |a-b| \) being even.
Step 1: Reflexive property.
For every element \(a \in A\), \[ |a-a| = 0 \] Since \(0\) is even, \( (a,a) \in R \) for all \(a \in A\). Hence, the relation \(R\) is reflexive.
Step 2: Symmetric property.
If \( (a,b) \in R \), then \[ |a-b| \text{ is even. \] But \[ |b-a| = |a-b| \] Thus \( |b-a| \) is also even, implying \( (b,a) \in R \). Hence, the relation \(R\) is symmetric.
Step 3: Transitive property.
Suppose \( (a,b) \in R \) and \( (b,c) \in R \). Then \[ |a-b| \text{ is even} \quad \text{and} \quad |b-c| \text{ is even}. \] This implies \[ a-b = 2m, \qquad b-c = 2n \] for some integers \(m,n\). Adding the equations, \[ a-c = 2m + 2n = 2(m+n) \] Thus \(a-c\) is even, which means \[ |a-c| \text{ is even} \] Therefore \( (a,c) \in R \). Hence, \(R\) is transitive.
Step 4: Conclusion.
Since the relation \(R\) satisfies reflexive, symmetric, and transitive properties, it is an equivalence relation on the set \(A\).