Question

Prove that in the set $\mathbb{Z}$ of integers the relation $R$ given by $R = \{(a,b): 5 \text{ divides } (a-b)\}$ is an equivalence relation.

Hint:

To prove equivalence: Always check reflexive, symmetric, transitive. Congruence modulo $n$ (like mod 5 here) is a standard equivalence relation.

Answer 1

A relation $R$ is an equivalence relation if it is reflexive, symmetric, and transitive. 1. Reflexive: For any $a \in \mathbb{Z}$, $a-a = 0$, and $5$ divides $0$. So, $(a,a) \in R$. Hence $R$ is reflexive. 2. Symmetric: Suppose $(a,b) \in R $\Rightarrow$ 5 | (a-b)$. Then $a-b = 5k$ for some $k \in \mathbb{Z}$. So $b-a = -5k$, which is also divisible by $5$. Hence, $(b,a) \in R$. So $R$ is symmetric. 3. Transitive: Suppose $(a,b) \in R$ and $(b,c) \in R$. Then $a-b = 5m$, $b-c = 5n$ for some $m,n \in \mathbb{Z}$. Adding: $(a-b) + (b-c) = a-c = 5(m+n)$. So, $5 | (a-c) $\Rightarrow$ (a,c) \in R$. Thus, $R$ is transitive. Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation. Final Answer: \[ \boxed{\; R \text{ is an equivalence relation.} \;} \]