Given a real number $x$ there exist unique numbers $n$ and $\epsilon$ such that $x=n+\epsilon$ where $n$ is an integer and $0\leq\epsilon<1$ 
Given a real number $x$ there exist unique numbers $n$ and $\epsilon$ such that $x=n+\epsilon$ where $n$ is an integer and $0\leq\epsilon<1$

I wrote: 
Since $0\leq\epsilon<1$ and $x=n+\epsilon\implies$ $\epsilon=x-n$, this shows that $n\leq x$.
Since $n$ is an integer we also know that $0\leq n \geq 1$.
Since $x = n + \epsilon \implies n = x - \epsilon$, and n is an integer, $x-\epsilon$ must be an integer. 
I don't think I'm close to a proof, how should I proceed? 
 A: There are two things to prove here, existence and uniqueness. Let's start with uniqueness.
Suppose $x = n + \epsilon = n' + \epsilon'$ where $n,n' \in \mathbb{Z}$ and $\epsilon,\epsilon' \in [0,1)$. Subtracting, we find $n-n' = \epsilon'-\epsilon$. Since $n-n' \in \mathbb{Z}$ we have $\epsilon-\epsilon' \in \mathbb{Z}$. But $|\epsilon-\epsilon'| < 1$, so $\epsilon-\epsilon' = 0$, i.e. $\epsilon = \epsilon'$. Thus $n+\epsilon = n' + \epsilon$ and so $n=n'$. Thus the expansion of $x=n+\epsilon$ is unique.
Now let's show existence. The statement is clearly true for $x \in [0,1)$, namely take $n=0$ and $\epsilon=x$. Now suppose that we know the theorem for $y \in [n-1,n)$.
Then for $x \in [n,n+1)$ we have $x-1 \in [n-1,n)$ so we can write $x-1 = n + \epsilon$. Then $x = (n+1)+\epsilon$ as required. A similar argument will show that if $x \in [n-2,n-1)$ the result holds as well. The result now follows by two inductions.
A: Hint:
$$x=\underbrace{\lfloor x\rfloor}_{=n}+\underbrace{\{x\}}_{=\varepsilon}.$$
