The relation $\sim$ in $\mathbb{R}$ is defined as: $x \sim y \iff x − y \in \mathbb{Z}$ $(a) \quad$ Show that $\sim$ is an equivalent relation.
$(b) \quad$ Give $2$ distinct equivalent classes (must show they are distinct).
$(c) \quad$ $[0, 1) = \{x \in \mathbb{R} \mid 0 \leq x < 1\}.$ Show that $^{\mathbb{R}}\big/_{\sim} \cong [0, 1).$
I was able to prove that the relation is reflexive, symmetric and transitive. So, it is equivalent.
For $(b)$ one class I could find was
$$\{y \in \mathbb{R} \mid x \sim y\} = \{y \in \mathbb{R} \mid \exists z \in \mathbb{Z}, x - y = z\} = \{y \in \mathbb{R} \mid \exists z \in \mathbb{Z}, x = z + y\}.$$
But I cannot find another one which is distinct!
And for $(c)$ I don't even know how to start this proof. Any ideas? Thanks

In your answer for b) you have to declare what $x$ is.
For instance if $x = \pi$ then we can define the equivalence class of pi is
$\{y\in \mathbb R: \pi\sim y\} = \{y\in \mathbb R|\exists z \in \mathbb Z; \pi-y = z\}=\{y\in \mathbb R| y=-z-\pi;z\in \mathbb Z\} = \{k-\pi|k\in \mathbb  Z\}=$
$\{..., -3-\pi, -2-\pi, -1, -\pi, 1-\pi, 2-\pi, 3-\pi, ....\}$.
You can do this for any real number $x$.
Do it for $x = \sqrt 2$ and get $\{y\in \mathbb R: \sqrt 2 \sim y\} = \{k-\sqrt 2|k\in \mathbb Z\} = \{...., -1 -\sqrt 2, -\sqrt 2, 1-\sqrt 2, 2-\sqrt 2,....\}$
Or eve for $x = 3$.  The equivalence class for $3$ is $\{k-3|k\in \mathbb Z\} = \mathbb Z$.
To show $\{..., -3-\pi, -2-\pi, -1, -\pi, 1-\pi, 2-\pi, 3-\pi, ....\}$ and  $\{...., -1 -\sqrt 2, -\sqrt 2, 1-\sqrt 2, 2-\sqrt 2,....\}$ are distinct, is a matter of saying
Let  $x \in \{y\in \mathbb R: \sqrt 2 \sim y\}\cap \{y\in \mathbb R: \pi\sim y\}$.  let $\pi \sim x$ and $\sqrt 2 \sim x$.  So $\pi -x = k\in \mathbb Z$ and $\sqrt 2-x= m\in \mathbb Z$ for some $k,m$.  But that means $x= \pi - k = \sqrt 2-m$ and that means $\pi = \sqrt 2 +(k -m)$ or that $\pi \sim 2$ (which follows by symmetry and transitivity).
And that's nonsense.  SO the two classes are distinct.
