$n,i,j \in\mathbb{N}$
$r \in \mathbb{R^{+}}$
$k = \left \lfloor{r}\right \rfloor $ (integer part of $r$)
$round ()$ = round function to make an integer
$round (n \times r) = ik + j(k+1)$
For example, Assume that $n = 8$, $r = 3.2$, then $round(8 \times 3.2) = round(25.6) = 26 = 2\times3 + 5\times(3+1)$.
how to prove this? or any related problems are welcome.
I will use floor and ceiling function notation.
Prove that $$n \lfloor r \rfloor \le \lfloor nr \rfloor \le \text{round}(nr) \le \lceil nr \rceil \le n \lceil r \rceil \le n (\lfloor r \rfloor + 1).$$
Let $k := \lfloor r \rfloor$. The natural numbers from $nk$ to $n(k+1)$ can be written as $$nk, (n-1)k + (k+1), (n-2)k + 2(k+1), \ldots, k + (n-1)(k+1), n(k+1).$$ Note that $\text{round}(nr)$ must be one of the above.
