<s>[INST] Is there any relation between $\lfloor n/k\rfloor$ and $\lfloor n/(k+1)\rfloor$? Given two integers $n$ and $k$ such that $n\geq k+1$. 
Can we find any relation between $\left\lfloor\dfrac{n}{k}\right\rfloor$ and $\left\lfloor \dfrac{n}{k+1}\right\rfloor$?
At first, I thought that $\left\lfloor \dfrac{n}{k+1}\right\rfloor=\left\lfloor\dfrac{n}{k}\right\rfloor-1,$ but then I found that, for $k=1$, $$\left\lfloor\dfrac{n}{2}\right\rfloor\neq n-1$$. So, I guess that $$\left\lfloor \dfrac{n}{k+1}\right\rfloor\leq\left\lfloor\dfrac{n}{k}\right\rfloor-1.$$ How can we prove this? Is this the best bound?
 [/INST] First let's consider that
$$
\eqalign{
  & \left\lfloor {x + y} \right\rfloor  =   \cr 
  &  = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor  =   \cr 
  &  = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left[ {1 - \left\{ x \right\} \le \left\{ y \right\}} \right] \cr} 
$$
where the curly brackets denotes the fractional part and where
in the last line $[P]$ denotes the Iverson bracket.
Then we have
$$
\eqalign{
  & \left\lfloor {{n \over k}} \right\rfloor  = \left\lfloor {{n \over {k + 1}}{{k + 1} \over k}} \right\rfloor
  = \left\lfloor {{n \over {k + 1}} + {n \over {k\left( {k + 1} \right)}}} \right\rfloor  =   \cr 
  &  = \left\lfloor {{n \over {k + 1}}} \right\rfloor  + \left\lfloor {{n \over {k\left( {k + 1} \right)}}} \right\rfloor  
+ \left[ {1 - \left\{ {{n \over {k + 1}}} \right\} \le \left\{ {{n \over {k\left( {k + 1} \right)}}} \right\}} \right] \cr} 
$$
which tells much about the difference between the two floors.
</s>