For $\displaystyle n \in \mathbb{N}$ show that $\displaystyle \left\lfloor \frac{x}{n} \right\rfloor = \left\lfloor \frac{\lfloor x \rfloor}{n} \right\rfloor$.

My proof...

If $\displaystyle n > x$ both sides will be 0 so we will deal with the case when $\displaystyle n <= x$.

Let $\displaystyle k \neq 0$ be the largest integer smaller than $\displaystyle x$ such that $\displaystyle n | k$.

Then we have $\displaystyle \lfloor x \rfloor = k + l$ where $\displaystyle l \in \mathbb{N}$ and $\displaystyle 0 \leq l \leq n-1$.

So $\displaystyle \left\lfloor \frac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor \frac{k + l}{n} \right\rfloor = \frac{k}{n} + \left\lfloor \frac{l}{n} \right\rfloor$ and since $\displaystyle \frac{l}{n} < 1$, $\displaystyle \frac{k}{n} + \left\lfloor \frac{l}{n} \right\rfloor = \frac{k}{n} + 0 = \frac{k}{n}$

Now let $\displaystyle x = \lfloor x \rfloor + \delta$ where $\displaystyle 0 \leq \delta < 1$.

Then $\displaystyle \left\lfloor \frac{x}{n} \right\rfloor = \left\lfloor \frac{k+l}{n} + \frac{\delta}{n} \right\rfloor = \frac{k}{n} + \left\lfloor \frac{l + \delta}{n} \right\rfloor$.

And since $\displaystyle 0 \leq l \leq n-1$, $\displaystyle l + \delta < n$, hence $\displaystyle \frac{l + \delta}{n} < 1 => \left\lfloor \frac{l + \delta}{n} \right\rfloor= 0$.

So $\displaystyle \frac{k}{n} + \left\lfloor \frac{l + \delta}{n} \right\rfloor = \frac{k}{n} + 0 = \frac{k}{n}$. Hence the two are equal.

I hope I've written this out right. To explain if it looks wrong... Imagine, 13.86/4. Then the largest integer smaller than $\displaystyle x$ that $\displaystyle n$ divides would be 12, $\displaystyle l$ would be 1 and $\displaystyle \delta$ would be 0.86.

EDIT: I suppose this proof would also work for the case for $\displaystyle x < n$ actually.