Show that
$\left \lfloor \dfrac{ \lfloor x \rfloor+m }{n} \right \rfloor = \left \lfloor \dfrac{x+m}{n} \right \rfloor$, where m is an arbitrary integer and $ n$ is an integer $> 0$.
Let us start with splitting up the fraction:
$$\left\lfloor\frac{\lfloor x\rfloor +m}{n}\right\rfloor=\left\lfloor\frac{\lfloor x\rfloor}{n}+\frac{m}{n}\right\rfloor$$ We have the equality $\lfloor x\rfloor = x-\{x\}$. Since $n$ and $m$ are constants, we can use this rule on the splitted fraction above, so that we can write:
$$\frac{\lfloor x\rfloor}{n}=\frac{x}{n}-\frac{\{x\}}{n},$$ Then inserting this in the top equality, we get:
$$\left\lfloor\frac{\lfloor x\rfloor +m}{n}\right\rfloor=\left\lfloor\frac{x}{n}+\frac{m-\{x\}}{n}\right\rfloor$$
Then merge the two fractions:
$$\left\lfloor\frac{\lfloor x\rfloor +m}{n}\right\rfloor=\left\lfloor\frac{x+m-\{x\}}{n}\right\rfloor$$
Since $\{x\}$ is a part of the fraction $\frac{m-\{x\}}{n}$, we can set the equality:
$$\frac{m-\{x\}}{n}=\left\lfloor\frac{m}{n}\right\rfloor,$$
and obtain:
$$\left\lfloor\frac{\lfloor x\rfloor +m}{n}\right\rfloor=\left\lfloor\frac{x+m}{n}\right\rfloor$$
However, I am not sure if the last step requires additional intermediary steps, and if this last step holds at all.
Any comments welcomed.
Thanks