Let $S=\{1,2,3,4,5,6,7,8\}$. Let $K$ be the set of all subsets of $S$ without the empty set. Let $\min X$ for $X\in K$ be the smallest number in $X$ and $\max X$ be the biggest number in $X$. Let $E$ be a relation over $K$.
For $A,B \in K$:
$(A,B)\in E \iff \min A+\max B=\min B + \max A$. Prove that $E$ is an equivalence relation and how many equivalence classes are in $E$?
To prove that $E$ is an equivalence relation is quite easy. The harder part is to find how many classes are. I think that for some $X\in K$, $|X| > 1$, $X$ is in relation with: $$ 2^{\max X-\min X - 1} $$ subsets of $K$. If $|X|=1$ then it's in relation with itself.
For example, $\{1, 5\}$ is in relation with: $$ \{1,5\}, \{1,2,5\}, \{1,3,5\}, \{1,4,5\}, \{1,2,3,5\}, \{1,2,4,5\}, \{1,3,4,5\}, \{1,2,3,4,5\} $$ that is with $2^{5-1-1}=8$ sets.
So the total numbers of equivalence classes is ${8 \choose 2}\cdot 2^{\max X-\min X -1}+8$ that is ${8 \choose 2}\cdot 2^{\max X-\min X -1}$ classes where $\max X\neq \min X$ and $8$ classes for every $X$ which is composed of only one number. But I can't arrive to the final answer without $\max, \min$ variables.
Also how can I prove that for $|X|>1$, $X$ is in relation with exactly $2^{\max X-\min X -1}$? It makes sense from manual calculations but what would be the formal proof? Is it because of the fact that for each $X$ $\max X$ and $\min X$ are unique?
Hint: Note that $$ \min A + \max B = \min B + \max A $$ is equivalent to $$ \max A -\min A = \max B - \min B $$ Now it ought to be a lot easier to tell that it is an equivalence relation, and what the equivalence classes are.