<s>[INST] Find the domain of $f(x)=\sec^{-1}\dfrac{x}{\sqrt{x-[x]}}$ 
Domain of $f(x)=\sec^{-1}\dfrac{x}{\sqrt{x-[x]}}$ is:

$$
x-[x]\neq0\implies x\notin \mathcal{Z}\implies x\in\mathcal{R}-\mathcal{Z}\\
\sec^{-1}:\mathcal{R}-(-1,1)\to(0,\pi)\implies \dfrac{x}{\sqrt{x-[x]}}\in\mathcal{R}-(-1,1)\\
\dfrac{x}{\sqrt{x-[x]}}\notin(-1,1)\\
x-[x]\in(0,1)\implies\sqrt{x-[x]}\in(0,1)
$$
The solution given in my reference is $\mathcal{R}-\{(-1,1)\cup\mathcal{Z}\}$, but how do I prove the part $x\notin (-1,1)$ ?
 [/INST] In order to take the square root (and have a real result) we need $x - \lfloor x \rfloor \ge 0$ , but that is not a problem.
We cannot divide by $0$
$x - \lfloor x \rfloor \ne 0 \implies x\notin \mathbb Z$
For inverse secant to be defined we need
$\left|\frac {x}{\sqrt{x - \lfloor x \rfloor}}\right|\ge 1$
$|x| \ge \sqrt{x - \lfloor x \rfloor}\\
x^2 \ge x - \lfloor x \rfloor\\
x^2 - x + \lfloor x \rfloor \ge 0$
We know  $x - 1 <\lfloor x \rfloor < x$
If $|x| > 1$ we are safe.
if $0<|x| < 1$ 
$x^2 - x + \lfloor x \rfloor = x^2 - x < 0$
if $-1<|x| < 0$ 
$x^2 - x - 1 = (x - \frac 12 + \frac {\sqrt 5}{2})(x - \frac 12 - \frac {\sqrt 5}{2})$
$x^2 - x - 1\le 0 \implies \frac 12 - \frac {\sqrt{5}}{2}\le x\le \frac 12 + \frac {\sqrt{5}}{2}$
What does that leave us with?
The domain is:
$x = \mathbb R - \mathbb Z - (0,1) - [\frac12 - \frac {\sqrt {5}}{2}, 0)$ 
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