Define a relation $\sim$ on $\Bbb R$ by $x\sim y\iff x-y\in\Bbb Z$. Show for any $x\in\Bbb{R}$, there is a unique $y\in [0,1)$ such that $x\sim y$. b) Show that for any $x \in \mathbb{R}$, there is a unique $y \in [0,1)$ such that $x \sim y$.
In part a), I've proven that $\sim$ is an equivalence relation.
I'm stuck on how to put my thoughts into a proof for b.
I've gotten that for any $x$, then $y$ would just be the decimal portion of $x$.
e.g. $x=3.123$ then $y=.123$ since $x-y=3.123-.123 \in \mathbb{Z}$
Basically my question is: Is it supposed to be $y=$ something and how do I turn that into a proof?
 A: For each $x \in \mathbb{R}$, we want to show that

*

*there exists

*a unique

$y \in [0, 1)$ such that $x \sim y$. We have to prove both assertions.
You have the right idea for the existence part. Subtract off the integer part. There's actually a notation for this, using the floor function:
$$
\lfloor x \rfloor = \max \{ n \in \mathbb{Z} \mid n \leq x \}.
$$
By the way, if you're trying to be rigorous, you have to show that the floor function is well-defined, i.e. that the set of integers has a maximum. Can you see how to do that?
A direct consequence of $N = \lfloor x \rfloor \in \mathbb{Z}$ is that
$$
N \leq x < N+1.
$$
Can you see how?
Now, we define $y$ to be the remainder:
$$
y = x - \lfloor x \rfloor = x - N. 
$$
The previous inequality directly implies that
$$
0 \leq y < 1, 
$$
as desired. This establishes existence.
For uniqueness, suppose that we have $y_1, y_2 \in [0, 1)$ such that $x \sim y_1$ and $x \sim y_2$. Since we have established that this is an equivalence relation, $y_1 \sim y_2$. Can you see how this implies that $y_1 = y_2$? Hence, the representative $y$ is unique.
A: Take $y\in\mathbb{R}$. Let $[y]$ be the inferior integer part of $y$. Then $y-[y]\in [0,1)$.
If $x\sim y$ and $x,y\in [0,1)$ then
$0\leq y-x=k< 1$
So we are looking for an integer $k$ less than $1$. Hence $k=0$ and $x=y$.
This answer to question $b)$.
Remark: Take $p\colon \mathbb{R}\to S^1$ sending $\alpha$ to $e^{2\pi i\alpha}$. Then you can observe that $p$ can be factorised by the quotient map $\mathbb{R}\to \mathbb{R}/\mathbb{Z}$. This provide us a map $p\colon \mathbb{R}/\mathbb{Z}\to S^1$. I leave you to prove that is an homeomorphism. Thus $ \mathbb{R}/\mathbb{Z}\cong S^1$.
This could be also another way to prove the statement $b)$.
A: Your idea is essentially correct. We can formalize it with the floor and fractional part functions:

*

*$\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\newcommand{\fp}[1]{\left\{ #1 \right\}}
\floor{x}$ is the largest integer $\le x$, i.e. $\floor{x} := \max \{ k \in \mathbb{Z} \mid k \le x \}$


*$\fp{x}$ is the fractional part, defined by $\fp{x} := x - \floor{x}$
Here, then, you want $y = \fp{x}$.
Correspondingly, $y \in [0,1)$ and $x - y = x - \fp{x} = x - (x-\floor{x})= \floor{x} \in \mathbb{Z}$.

If you don't want to appeal to alternative functions, do so with decimal expansions. Write
$$x := u\sum_{i \in \mathbb{Z}} d_i 10^i = \pm \cdots d_3d_2d_1d_0 . d_{-1}d_{-2}d_{-3} \cdots$$
where $d_i = 0$ for all sufficiently large $i$, and $d_i \in \{0,1,2,\cdots,9\}$ generally. Finally, $u \in \{+1,-1\}$, in order to account for the sign of $x$.
Then the $y$ you want is
$$y := \begin{cases}
\displaystyle \sum_{i = -\infty}^{-1} d_i 10^i =   0.d_{-1}d_{-2}d_{-3} \cdots & x \ge 0 \\
1 - \displaystyle \sum_{i = -\infty}^{-1} d_i 10^i & x < 0
\end{cases}
$$

Uniqueness in either framework I leave up to you, but it's fairly straightforward.
