1. ## Floor functions!

Given any real number $x$ and positive integer $n$, prove that $\left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{x}{n}\right \rfloor$

Deduce that for any real number $y$ and positive integers $n$ and $m$, one has

$\left \lfloor\frac{\left \lfloor \frac{y}{m}\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{\left \lfloor \frac{y}{n}\right \rfloor}{m}\right \rfloor$

I think the second part is very straightforward i.e. $\left \lfloor\frac{\left \lfloor \frac{y}{m}\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{\frac{y}{m}}{n}\right \rfloor = \left \lfloor\frac{\frac{y}{n}}{m}\right \rfloor = \left \lfloor\frac{\left \lfloor \frac{y}{n}\right \rfloor}{m}\right \rfloor$

I just can't prove the initial part.

Any help appreciated!

2. ## Re: Floor functions!

Let $x=qn+r+a$ where $0\le r and $0\le a<1$ and q and r are integers.

I think this goes somewhere.

3. ## Re: Floor functions!

Cheers, think I get it.. So by putting $x=qn+r+a$ as you suggest we have $\left \lfloor x\right \rfloor = qn+r$ and as $r so $\left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = q$

Now $\frac{x}{n} = q+ \frac{r+a}{n}$ as $r\in Z$ and $r, and $a<1$ we have $(r+a) so $\left \lfloor q+ \frac{r+a}{n}\right \rfloor = q$ also.

Therefore $\left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{x}{n}\right \rfloor$