Proof Verification: Show that $[mx] = \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg]$ I was wondering if this proof is valid. 
I use $[x]$ to denote the floor of $x$.
Problem 
Prove that
$$[mx] = \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg]$$
where $m \in \mathbb{N}$ and $x \in \mathbb{R}$.
Proof
Let $m \in \mathbb{N}$ and let $x \in \mathbb{R}$ such that $\epsilon = x - [x]$ where $0 \leq \epsilon < 1$.
Partition the interval $[0,1)$ as 
$$[0,1) = \bigcup_{k=0}^{m-1} \, \bigg[\frac{k}{m}, \frac{k+1}{m} \bigg)$$
Let $p \in \{1, \dotsc, m\}$ and consider the interval
$$\frac{p-1}{m} \leq \epsilon < \frac{p}{m}$$
Expanding  and simplifying the sum $\sum_{k=0}^{m-1} \, \big[x+\frac{k}{m} \big]$ renders
$$\begin{align} \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg] &= [x] + \bigg([x] +  \bigg[\epsilon + \frac{1}{m}\bigg]\bigg) + \cdots + \bigg([x] + \bigg[\epsilon + \frac{m-1}{m} \bigg]\bigg) \\ &= m[x] + \sum_{k=1}^{m-1} \, \bigg[\epsilon + \frac{k}{m} \bigg] \end{align} $$
Since each term in the sum $\sum_{k=1}^{m-1} \, \big[\epsilon + \frac{k}{m} \big]$ either equals 0 or 1 depending on $p$, we observe that there are at most $(m-p)$ zeros and $(p-1)$ ones amongst the $(m-1)$ terms of the given series. Hence,
$$\begin{align} \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg] &= m[x] + \sum_{k=1}^{m-1} \, \bigg[\epsilon + \frac{k}{m} \bigg] \\\\ &= m[x] + (m-p)\cdot 0 + (p-1) \cdot 1 \\\\ &= m[x] + (p-1) \end{align}$$
Furthermore, $\frac{p-1}{m} \leq \epsilon < \frac{p}{m}$ implies $p-1 \leq m \cdot \epsilon < p$. Thus we can write
$$\begin{align} [mx] &= \big[m([x]+\epsilon)\big] \\\\ &= m[x] + [m \cdot \epsilon] \\\\ &= m[x] + (p-1). \end{align}$$
Since for all intervals $p = 1, \dotsc, m$ we have 
$$\begin{align} \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg] &= [mx] \\ &= m[x] + (p-1) \end{align}$$
we conclude that $[mx] = \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg]$
 A: There are a couple of small bugs, but the argument is basically correct. When you partition $[0,1)$, you want either $\bigcup_{k=0}^{m-1}\left[\frac{k}m,\frac{k+1}m\right)$ or $\bigcup_{k=1}^m\left[\frac{k-1}m,\frac{k}m\right)$. In the next sentence you should not be considering all $m$ values of $p$: you already have $\epsilon=x-\lfloor x\rfloor$, and you’re defining $p$ to be the unique member of $\{1,\ldots,m\}$ such that 
$$\frac{p-1}m\le\epsilon<\frac{p}m\;.$$
A bit later you say that $\sum_{k=1}^{m-1}\left\lfloor\epsilon+\frac{k}m\right\rfloor$ is $0$ or $1$ depending on $p$; what you mean, I think, is that each term of that sum is $0$ or $1$.
The same basic idea can be carried out a bit more compactly.
Let $\ell\in\{0,\ldots,m-1\}$ be maximal such that $x+\frac{\ell}m<\lfloor x\rfloor+1$. Then $\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor$ for $k=0,\ldots,\ell$, and $\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor+1$ for $k=\ell+1,\ldots,m-1$. Thus,
$$\begin{align*}
\sum_{k=0}^{m-1}\left\lfloor x+\frac{k}m\right\rfloor&=(\ell+1)\lfloor x\rfloor+(m-1-\ell)(\lfloor x\rfloor+1)\\
&=m\lfloor x\rfloor+m-1-\ell\;.
\end{align*}\tag{1}$$
This also implies that
$$1-\frac{\ell+1}m\le x-\lfloor x\rfloor<1-\frac{\ell}m$$
and hence that
$$m-\ell-1\le mx-m\lfloor x\rfloor<m-\ell\;,$$
or
$$m\lfloor x\rfloor+m-\ell-1\le mx<m\lfloor x\rfloor+m-\ell\;.\tag{2}$$
But $(2)$ implies that $\lfloor mx\rfloor=m\lfloor x\rfloor+m-\ell-1$, which, when combined with $(1)$, yields the desired result:
$$\lfloor mx\rfloor=\sum_{k=0}^{m-1}\left\lfloor x+\frac{k}m\right\rfloor\;.$$
A: I think it's a good effort but your proof is difficult to follow due to some issues with presentation and definition.  For example, the partition of $[0,1) = \bigcup_{k=1}^m [k/m, (k+1)/m)$ is problematic because it obviously fails to contain $0$.  Also, if you say "let $p = 1, 2, \ldots, m$ and consider the $p^{\rm th}$ subinterval..." this is not clear, nor is it clear that if $p$ is to take on a sequence of values, that in fact the following inequality is true for at most one particular $p$.  At that point, I stopped reading carefully.
A more elegant proof of the stated identity exploits periodicity:  consider the function defined for all nonnegative integers $m$ and reals $x$: $$f_m(x) = -\lfloor mx \rfloor + \sum_{k=0}^{m-1} \left\lfloor x + \frac{k}{m} \right\rfloor.$$  Then note that $$\begin{align*} f_m(x+1/m) &= -\lfloor m(x+1/m) \rfloor + \sum_{k=0}^{m-1} \left\lfloor x + \frac{k+1}{m} \right\rfloor \\ &= -\lfloor mx \rfloor - 1+ \sum_{k=1}^m \left\lfloor x + \frac{k}{m} \right\rfloor \\ &=  f_m(x) - 1 + \lfloor x+m/m\rfloor - \lfloor x \rfloor \\ &= f_m(x).\end{align*}$$  That is to say, $f_m$ is a periodic function for which $1/m$ is an integer multiple of its period.  So it suffices to consider the behavior of $f_m(x)$ on the interval $x \in [0, 1/m)$.  But on this interval, $\lfloor mx \rfloor = 0$, and because $0 \le x + k/m < 1$ for all $0 \le k \le m-1$ and $0 \le x < 1/m$, all the terms in the sum are also zero; thus $f_m(x)$ is identically zero on $[0,1/m)$, and by extension, is zero for all $x \in \mathbb R$.  This completes the proof.
