Question

Prove that the relation $R = \{(a,b) \in \mathbb{Z} \times \mathbb{Z} \;|\; (a-b)$ is divisible by $2\}$ is an equivalence relation.

Hint:

To prove equivalence relation: check reflexive, symmetric, and transitive properties.

Answer 1

We need to prove that $R$ is reflexive, symmetric, and transitive.

Step 1: Reflexivity.
For any $a \in \mathbb{Z}$, we have $a-a = 0$, which is divisible by $2$. So, $(a,a) \in R$. Hence, $R$ is reflexive.

Step 2: Symmetry.
Suppose $(a,b) \in R$. Then, $(a-b)$ is divisible by $2$. So, $a-b = 2k$ for some $k \in \mathbb{Z}$. $\Rightarrow b-a = -2k$, which is also divisible by $2$. Thus, $(b,a) \in R$. Hence, $R$ is symmetric.

Step 3: Transitivity.
Suppose $(a,b) \in R$ and $(b,c) \in R$. Then, $a-b = 2k_1$, and $b-c = 2k_2$ for some integers $k_1, k_2$. Adding, $a-c = (a-b) + (b-c) = 2k_1 + 2k_2 = 2(k_1+k_2)$. Thus, $a-c$ is divisible by $2$ $\Rightarrow (a,c) \in R$. Hence, $R$ is transitive.

Step 4: Conclusion.
Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation.

Final Answer: \[ \boxed{R \text{ is an equivalence relation.}} \]