prove that $x∉ℤ⇒\lfloor{-x}\rfloor=-\lfloor{x}\rfloor-1$

Question

prove that $x∉ℤ⇒\lfloor{-x}\rfloor=-\lfloor{x}\rfloor-1$

So, I have that $(\lfloor{-x}\rfloor=n)⇔(n≤-x<n+1)$

And I also have that $(-\lfloor{x}\rfloor-1=n)⇔(\lfloor{x}\rfloor=-n-1)⇔(-n-1≤x<-n)$

I will try to get from the first definition to the second one:

$(\lfloor{-x}\rfloor=n)⇔(n≤-x<n+1)$

$⇔(-n≥x>-n-1)$

$⇔(-n-1<x≤-n)$

As you can see, the only thing is left to do, is to turn $(-n-1<x≤-n)$ into $(-n-1≤x<-n)$, nonetheless, I haven't been able to. How do you think I can complete the proof? Thanks in advance.

Answer 1

You know that $-n, -n -1 \in \mathbb Z$ while it is the hypothesis that $x \notin \mathbb Z$. Thus $x \neq -n, x \neq -n-1$. Thus we have $$(-n-1 < x \leq -n) \Leftrightarrow (-n-1 < x < -n)\Leftrightarrow (-n-1 \leq x < -n)$$