I have proof/idea but just wanted to get any extra advice.
Prove that given any $k \in \mathbb{N}$, the equivalence relation defined by $a \sim b \iff a \equiv b$(mod k) yields equivalence classes whose cardinalities are all the same. What is this cardinalitiy?
Just to see where to start with my proof, I just started with an arbitrary class of [2] just to see where to go. Then,
$[2] = \{x \in \mathbb{N} | k|2 -x\}$. Since $2 - x = k * z$ for some $z \in \mathbb{k}$. Since the natural numbers would be closed under the integers $[2] = \mathbb{N}$.
I was thinking about generalizing the equation. Let $r \in \mathbb{N}$. Then, $[r] = \{x \in \mathbb{N}|k|r - x\} = \mathbb{N}$.
My main issue/question is this seems like an example of one equivalence class with an infinite cardinality, as opposed to multiple equivalence classes with the same cardinality. Any help/advice would be appreciated!
Of course, the relation $a \sim b \iff a \equiv\ b\ (\mod k)$ produces $k$ classes, each with infinite cardinality. Members of any two classes can be placed in a one-to-one relationship with any other.
If there's anything more being asked here, I don't see it.