I need to compute the range of: $$f(x) = \frac{1}{1+\lfloor x \rfloor}$$
Intuitively, the range is $\frac{1}{n}$, where $n \in \Bbb Z^*$, but I want to prove that this is the range of the function algebraically, using the fact that:
$\lfloor x \rfloor = m \iff m \le x \lt m+1 $
and $y= \frac{1}{1+\lfloor x \rfloor} \iff \forall \lfloor x \rfloor \ne -1 \implies \lfloor x \rfloor = \frac {1 - y}{y} $
$\iff \frac {1 - y}{y} \le x \lt \frac {1 - y}{y} +1 , \forall y \ne 0$
But I don't know how to proceed.
Any hints?
From the step $\lfloor x \rfloor =\dfrac{1-y}{y}$, you get that $$\frac{1-y}{y} \in \Bbb{Z}-\{-1\}.$$ Thus $\dfrac{1-y}{y}=k$, where $k \neq -1$. So $y=\frac{1}{k+1}$, where $k \neq -1$.
NOTE: Although I have to say that the above is just rehashing the obvious in a technical manner.