# 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

$$x+\frac{n-1}{n}-1 \lt \left [x +\frac{n-1}{n}\right] \le x+\frac{n-1}{n} \tag{n}$$

$$nx+\frac{n(n-3)}{2} \lt S \le nx+\frac{n(n-1)}{2}$$

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 $\left [nx \right]$?

• The last sentence is simply not true. – Cave Johnson Apr 26 '18 at 15:17

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.
Adding all of them we get $$nx+\frac{\frac{n(n-1)}2}{n}-n<S\le nx+\frac{\frac{n(n-1)}2}{n}$$ or $$nx-\frac{n+1}2<S\le nx+\frac{n-1}2\implies -\frac n2<S-nx+\frac12\le\frac n2$$ and this interval gets wider and wider as $n$ gets larger.