Property of the floor function Given $u,v \in \mathbb{R}_{+}$, and let $n:=\lfloor v \rfloor$. where $u\in [0,1]$.
Is $\lfloor vu \rfloor =\lfloor nu \rfloor$ or $\lfloor vu \rfloor =\lfloor nu \rfloor+1$?
Edit: I missed the assumption that $u\in [0,1]$. 
AugSB gave counterexamples, even a case where $\lfloor vu \rfloor =\lfloor nu \rfloor +2$.
However can i bound $\lfloor nu \rfloor$ somehow, like
$\lfloor vu \rfloor \in [\lfloor nu \rfloor ,\lfloor nu \rfloor +k]$, i hope $k=2$ should be enough.

The equality does not hold in every case. Take, for instance, $v=7/2$ and try both for $u=3/10$ and $u=2/5$.
However you can bound $\lfloor u v\rfloor$ in the interval $\big[\lfloor u n\rfloor, \lfloor u n\rfloor+1 \big]$, where $n=\lfloor v\rfloor$. Let $\{x\}=x-\lfloor x\rfloor$ be the fractional part of $x$. Then:
$$\lfloor u v\rfloor = uv - \{uv\}$$
$$\lfloor u n\rfloor = un - \{un\} = uv - u\{v\}-\{un\}$$
Therefore, it is enough to show that:
$$ u\{v\}+\{un\}-1 \le \{uv\} \le u\{v\}+\{un\}$$
For $u=0$ and $u=1$ it is clearly true, so let us assume that $u\in(0,1)$. Moreover, if $v$ is integer, the inequality holds too. Thus, let us assume $z\not\in\mathbb{Z}$. Then
$$ u\{v\}+\{un\}-1 \le \{u\{v\}+ un\} \le u\{v\}+\{un\}. $$
Thus it reduces to show that
$$ a+\{b\}-1 \le \{a+ b\} \le a+\{b\}, $$
where $a\in(0,1)$ and $b>0$. Try to follow from here.
