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


1 Answer 1


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

  • $\begingroup$ 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 $$n=\text{floor}\left(\frac{L-1}{k}\right) \Leftrightarrow n \leq \frac{L-1}{k}<n+1$$ $\endgroup$ Commented Sep 17, 2018 at 7:27
  • $\begingroup$ @K. Claesson: That's the defining property of the floor-function. $\endgroup$ Commented Sep 17, 2018 at 7:32
  • $\begingroup$ For $x \in \mathbb{R}$ you have $$floor(x) = [x]= max\left\{ n \in \mathbb{Z}: n \leq x \right\}$$ $\endgroup$ Commented Sep 17, 2018 at 7:40
  • $\begingroup$ 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. $\endgroup$ Commented Sep 17, 2018 at 7:49

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