We define eqiuvalence relation $\equiv$ at set $P(\mathbb{Z})$ as follow:
$A \equiv B \iff |A \setminus \mathbb{N}| = |B \setminus\mathbb{N}|$
a)find equivalence class for $\emptyset$
b)find cardinality of quotient set $P(\mathbb{Z})$
c)find cardinality of equivalence classes
my attempt:
a) $[\emptyset]= \{B \in P(\mathbb{N}) \}$ since then $|B \setminus\mathbb{N}|= \emptyset$ (I don't know if the way I described this set is proper but the idea is such set contains only natural numbers)
b)if $A,B\in P(\mathbb{Z\setminus N})$ and $A \neq B$ then $A \not\equiv B$ and since $|P(\mathbb{Z\setminus N})|=|\mathbb{R}|$ then our set is continuum
c) if $A$ has only finite amount of negative integers say $k$ then $|[A]|=\aleph_0$ since we're picking $k$ negative integers from $\mathbb{Z}$ and finite union of countable is countable
if $A$ has infintely many negative integers then we have $|[A]|=\aleph_0^{\aleph_0}=\aleph_1$ ( I have problem how to explain that but that's my intuition)
if $A$ has no negative integers then I think also $\aleph_1$ since at this equiv class will be $P(\mathbb{N})$ whose cardinality is $\aleph_1$