# Floor functions!

• Apr 12th 2012, 08:05 AM
leshields
Floor functions!
Given any real number $\displaystyle x$ and positive integer $\displaystyle n$, prove that $\displaystyle \left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{x}{n}\right \rfloor$

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

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

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

I just can't prove the initial part.

Any help appreciated!
• Apr 12th 2012, 09:07 AM
a tutor
Re: Floor functions!
Let $\displaystyle x=qn+r+a$ where $\displaystyle 0\le r<n$ and $\displaystyle 0\le a<1$ and q and r are integers.

I think this goes somewhere.
• Apr 13th 2012, 10:27 AM
leshields
Re: Floor functions!
Cheers, think I get it.. So by putting $\displaystyle x=qn+r+a$ as you suggest we have $\displaystyle \left \lfloor x\right \rfloor = qn+r$ and as $\displaystyle r<n, \left \lfloor \frac{r}{n}\right \rfloor = 0$ so $\displaystyle \left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = q$

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

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