<s>[INST] Proving Hermite's Identity in a different approach To prove 
$$S=\left [x \right]+\left [x+\frac{1}{n} \right]+\left [x +\frac{2}{n}\right]+\cdots+\left [x +\frac{n-1}{n}\right]=\left [nx \right]$$
using and starting with $$x-1 \lt \left [x \right]\le x \tag{1}$$ we have
$$x+\frac{1}{n}-1 \lt \left [x +\frac{1}{n}\right] \le x+\frac{1}{n} \tag{2}$$
$$x+\frac{2}{n}-1 \lt \left [x +\frac{2}{n}\right] \le x+\frac{2}{n} \tag{3}$$
$$x+\frac{3}{n}-1 \lt \left [x +\frac{3}{n}\right] \le x+\frac{3}{n} \tag{4}$$
and so on till
$$x+\frac{n-1}{n}-1 \lt \left [x +\frac{n-1}{n}\right] \le x+\frac{n-1}{n} \tag{n}$$
Adding all we get
$$nx+\frac{n(n-3)}{2} \lt S \le nx+\frac{n(n-1)}{2}$$
Now how can we prove that between $nx+\frac{n(n-3)}{2}$ and $nx+\frac{n(n-1)}{2}$ there is only on integer which is $\left [nx \right]$?
 [/INST] Here is a proof I came up with
about a year ago:
For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$
The basic idea is that
you have to see
how far $x$ is from
$\lfloor x \rfloor$
and,
if $d = x-\lfloor x \rfloor$,
what $\lfloor nd \rfloor$ is.
This allows you do get the sum.
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