Let $x \in \Bbb R$ and $n \in \Bbb Z$ with $n \gt 0$. Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$;
in particular, $\left\lfloor\dfrac{\left\lfloor\dfrac {a}{b} \right\rfloor}{c} \right\rfloor = \left\lfloor\dfrac{a}{bc}\right\rfloor$
I don't know how to approach this problem, any hints to solve this would be greatly appreciated.
Thanks for your help!
Let $m=\lfloor x\rfloor$. Then $m\leq x < m+1$. So $\dfrac{m}{n}\leq \dfrac{x}{n}< \dfrac{m+1}{n}$. By the division algorithm, $m =qn+r$ for some integers $q,r$ with $0\leq r\leq n-1$. Thus, $$q\leq\dfrac{m}{n}\leq\dfrac{x}{n}<\dfrac{m+1}{n}=\dfrac{qn+r+1}{n}\leq q+1.$$ So $\lfloor \tfrac{m}{n} \rfloor=\lfloor\tfrac{x}{n}\rfloor=q$.
Note: $\lfloor x\rfloor$ is the unique integer $N$ such that $N\leq x<N+1$.
A much more general result is available in Donald Knuth's Concrete Mathematics and it is attributed to "R. J. McEliece when he was an undergrad". He proved it for ceiling functions, but I included a proof for the floor function in a book that I am writing. Here is a variation of what I wrote in my book:
Let $f$ be a continuous and monotonically increasing function, defined on an interval $I$ in $\mathbb{R},$ with the property that $$f(x)\in\mathbb{Z}\implies x\in\mathbb{Z}.$$ Then, for all $x\in I,$ $$ \lfloor f(x) \rfloor =\left\lfloor f(\lfloor x\rfloor) \right\rfloor \text{ and } \lceil f(x) \rceil =\left\lceil f(\lceil x\rceil) \right\rceil .$$ Note that each identity holds only if the expression on either side of it is defined, as $x$ being in $I$ does not necessarily mean that $\lfloor x\rfloor$ and $\lceil x\rceil$ are in $I.$
I'll leave it to you to check out Knuth's ceiling proof and adapt it to the floor scenario. Do you see why this applies to your problem? Roots, logs and linear functions with positive slope all work here (linear in particular).
