What is the set \frac{\mathbb{Z}}{2\mathbb{Z}}?

Step 1
Fix ${\displaystyle n\in \mathbb{N}}$
The set
${\displaystyle \frac{\mathbb{Z}}{n\mathbb{Z}}=\{x\in \mathbb{Z}\mid x\mathrm{\neg }\{\in \}n\mathbb{Z}\}}$
where
${\displaystyle n\mathbb{Z}=\{ny\in \mathbb{Z}\mid y\in \mathbb{Z}\}}$
The group
${\displaystyle \frac{\mathbb{Z}}{n\mathbb{Z}}=\{x+n\mathbb{Z}\mid x\in \mathbb{Z}\}}$
where ${\displaystyle n\mathbb{Z}}$ is as above, is uderstood as a group under the operation
${\displaystyle (a+n\mathbb{Z})+(b+n\mathbb{Z})+=(a+b)+n\mathbb{Z}}$
It is a quotient group. Each of its elements is what is known as a coset of ${\displaystyle n\mathbb{Z}}$
Your case, of course, is when ${\displaystyle n=2}$

Step 1
${\displaystyle \mathbb{Z}}$ be the set of all integers.
We want to put all integers into n distinct buckets and two integers belong to same bucket if they are similar.
Now, it is meaningless until we define what does it mean by the word ''similar '' and ''bucket''.
Two integers a and b are similar ${\displaystyle (a\sim b)}$ if ${\displaystyle a-b}$ is divisible by n.
It is easy to check that the relation is an equivalence relation.
Then the equivalence class(bucket) of a,
${\displaystyle \left[a\right]=\{b\in \mathbb{Z}:b\sim a\}}$
${\displaystyle a\in \mathbb{Z}}$ and n>1 by division algorithm,
${\displaystyle a=nq+r}$ (q: quotient, r: reminder, ${\displaystyle 0\le r\le (n-1)}$)
${\displaystyle a-r=nq}$ and hence, ${\displaystyle a\sim r}$
${\displaystyle \Rightarrow \left[a\right]=\left[r\right]}$
Hence, there is n distinct equivalence classes
${\displaystyle \left[0\right],\left[1\right],\cdots ,[n-1]}$
Then the set of all equivalence classes
${\displaystyle \{\left[0\right],\left[1\right],\cdots ,[n-1]\}=\frac{\mathbb{Z}}{n\mathbb{Z}}}$
In your question, ${\displaystyle n=2}$
${\displaystyle \frac{\mathbb{Z}}{2\mathbb{Z}}=\left\{\begin{array}{c}0\\ 1\end{array}\right\}}$
Two integers belongs to same equivalence class iff they leave the same reminder upon division by 2.
Like, ${\displaystyle 2\in \left[0\right]}$ because both are divisible by 2.