# Is it possible to solve $n=\text{floor}\left(\frac{L-1}{k}\right)$ for $L$?

Is it possible to solve $$n=\text{floor}\left(\frac{L-1}{k}\right), n,k,L \in \mathbb{Z}^+$$ for $L$?

## 1 Answer

Hint: $$n \leq \frac{L-1}{k}<n+1$$

• Reasoning by analogy, I assume that if I can rewrite $n \leq \frac{L-1}{k}<n+1$ to be $L \leq \text{[something]} < L + 1$ that will be equivalent to saying that $L = \text{floor}[\text{something}]$, but how do you conclude that $$n=\text{floor}\left(\frac{L-1}{k}\right) \Leftrightarrow n \leq \frac{L-1}{k}<n+1$$ Commented Sep 17, 2018 at 7:27
• @K. Claesson: That's the defining property of the floor-function. Commented Sep 17, 2018 at 7:32
• For $x \in \mathbb{R}$ you have $$floor(x) = [x]= max\left\{ n \in \mathbb{Z}: n \leq x \right\}$$ Commented Sep 17, 2018 at 7:40
• I understand it now. Actually, it is really obvious. It helped me to think of why the inequality is true for a special case, e.g. floor(2.7) and then I could see how it applies generally. Commented Sep 17, 2018 at 7:49