If $n$ is an integer and $t$ any real number, then it is straightforward to show that$\lfloor t + n \rfloor = \lfloor t \rfloor + n.$
Therefore, since $\lfloor x \rfloor$ is an integer,
\begin{align*}
\quad &\lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor \\
= &\lfloor y + x \rfloor - \lfloor x \rfloor + \lfloor x \rfloor \\
= &\lfloor x + y \rfloor.
\end{align*}
Let $f(x)=\dfrac 1{\sqrt{\{x\}}}$ where $\{x\}$ designates the fractional part, it belongs to $[0,1)$.
This comes from the definition of integer part (LHS below): $$\lfloor x\rfloor\le x<\lfloor x\rfloor+1\implies 0\le\{x\}=x-\lfloor x\rfloor<1$$
In particular $\{x\}\ge 0$, even for negative numbers, so $\sqrt{\{x\}}$ is defined everywhere.
First of all notice that your function is $1$-periodic so you can study it on $[0,1]$.
Indeed $\{x+1\}=x+1-\lfloor x+1\rfloor=x+1-(\lfloor x\rfloor+1)=x-\lfloor x\rfloor=\{x\}$ so $f$ is $1$-periodic as well.
As you noticed $\{x\}=0$ whenever $x$ is an integer, so the local domain is $(0,1)$ and the global domain extended by periodicity is $\mathbb R\setminus\mathbb Z$.
For the range since on $(0,1)$ we have $0<\{x\}<1$ then we get $f(x)>1$ and the range is thus $(1,+\infty)$.
Best Answer
If $x,y\not\in\mathbb{Z}$ and $x+y\in\mathbb{Z}$, then $\left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor=x+y-1$. Therefore, $$ \left\lfloor j\frac ab\right\rfloor+\left\lfloor(b-j)\frac ab\right\rfloor=a-1\tag{1} $$ If $(a,b)=1$, then for $0\lt j\lt b$, we have $j\frac ab\not\in\mathbb{Z}$.
Summing $(1)$, we get $$ \sum_{j=1}^{b-1}\left\lfloor j\frac ab\right\rfloor+\left\lfloor(b-j)\frac ab\right\rfloor=(a-1)(b-1)\tag{2} $$ $(2)$ counts each term in the sum we want twice, so we get $$ \sum_{j=1}^{b-1}\left\lfloor j\frac ab\right\rfloor=\frac{(a-1)(b-1)}2\tag{3} $$