Mathematical Reflections 2014, Issue 1 - Problem J289

Problem:
Let $a$ be a real number such that $0 \leq a < 1$. Prove that
$$\left\lfloor a\left(1+\left\lfloor \dfrac{1}{1-a} \right\rfloor \right)\right\rfloor+1=\left\lfloor \dfrac{1}{1-a} \right\rfloor.$$

Proposed by Arkady Alt.

Solution:

Let $m \in \mathbb{N}$ and $\alpha \in [0,1)$ such that $\dfrac{1}{1-a}=m+\alpha$. Then, we have to prove that
$$\left\lfloor \left(1-\dfrac{1}{m+\alpha}\right)(1+m) \right\rfloor+1=m.$$
Indeed, $$\left\lfloor \left(1-\dfrac{1}{m+\alpha}\right)(1+m) \right\rfloor=\left\lfloor m+1-\dfrac{m+1}{m+\alpha} \right\rfloor=m-1,$$ where the last equality follows from $-2<-\dfrac{m+1}{m+\alpha}<-1$.