I am currently trying to understand the proof that $[-x] = -[x]-1$ if $x$ is not an integer (solutions for $b$.)
where $[x]$ is the floor function.
Can somebody explain me how he went from $$-n-1 < -x < n$$ to $$[-x] = -n - 1 = -[x]-1$$ in the solutions for $b$.
On
Another proof that may help:
Let $x = a + \epsilon$ where $a \in \Bbb Z, \epsilon \in (-1,1)$. This is a way to formulate $x$ in terms of its integer and decimal component.
Then $[x]$ = $a + [\epsilon]$ (try to convince yourself of this) and $[-x] = -a + [-\epsilon]$.
If $\epsilon = 0$ (ie $x$ is an integer and, thus, $[\epsilon] = 0$), then the $[-x] = -[x]$ equality should be obvious.
Suppose $\epsilon \gt 0$, then we have $[x] = a + 0 = a$ and $[-x] = -a + (-1)$. Then we have $[-x] = [x] - 1$.
If $\epsilon \lt 0$, then we have $[x] = a - 1$ and $ [-x] = -a + 0$. Again, $[-x] = -[x] -1$.
Since $-n-1<-x<-n$, the number $-n-1$ is the greatest integer less or equal $-x$ (it is less than $-x$ but the next integer, namely $(-n-1)+1=-n$ is greater).
Therefore $[-x]=-n-1$. Since $[x]=n$, we get $[-x]=-n-1=-[x]-1$.