I am having trouble with the following proof:
Define the relation $R$ on $\mathbb{Z}$ by $nRm$ if $n-m$ is divisible by $2$. Prove that the equivalence class for $0^{(\bar{0})}$ and the equivalence class for $1^{(\bar{1})}$ are disjoint.
I completely understand how to prove that some relation is an equivalence relation, but I'm still a little unclear regarding the idea of an equivalence class. We had our last day of classes on Friday and my teacher went over equivalence classes fairly quickly, and my notes don't make much sense.
I know that $0^{(\bar{0})}$ is the set of all even numbers and that $1^{(\bar{1})}$ is the set off all odd numbers.
So then $\bar{0}$ is the equivalence class for $0^{(\bar{0})}$ and $\bar{1}$ is the equivalence class for $1^{(\bar{1})}$? I know that $\bar{0} \cup 1\bar{1}=\mathbb{Z}$ and that I must show that $\bar{0} \cap \bar{1}=\emptyset$. But I am confused how I would show that by contradiction. If I assume $\bar{0} \cap \bar{1}\neq\emptyset$ and let $x\in\bar{0} \cap \bar{1}$, how do I go from there?
Really, the only definition I have for an equivalence class for $R$ on $X$ is if $x\in X$ then $\bar{x}=\left\{ y\in X: xRy\right\}$.
Hint: if $x\in\bar{0}\cap \bar{1}$, then in particular $x\in\bar{0}$ and $x\in\bar{1}$. Now, $x\in\bar{0}$ means that $xR0$, so that $x-0=x$ is divisible by $2$ so that $x$ is even. Perform a similar computation assuming that $x\in\bar{1}$ and derive a contradiction.