How can find domain and range of $f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor}$? How can find $D_{f(x)},R_{f(x)}$ when $$f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor} \ ?$$
I help some problem like this for a question here ... (Domain and range of a floor function)
But I get stuck on this to find Domain and Range of above function .
If You can help me to find out , I will be very thankful .
 A: So, with the standard notation for the  floor function, we are to study
$$ \bbox[lightyellow] {  
y = \sqrt {\sqrt 2 \left\lfloor x \right\rfloor  - \left\lfloor {\sqrt 2 x} \right\rfloor } 
} \tag {1}$$
Let's first introduce some additional notation concerning the fractional part $\{x\}$
$$
x = \left\lfloor x \right\rfloor  + \left\{ x \right\}\quad 0 \le \left\{ x \right\} < 1
$$
and the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$
Then the radicand can be written as
$$ \bbox[lightyellow] {  
\eqalign{
  & r = \sqrt 2 \left\lfloor x \right\rfloor  - \left\lfloor {\sqrt 2 x} \right\rfloor  =   \cr 
  &  = \sqrt 2 \left( {x - \left\{ x \right\}} \right) - \left( {\sqrt 2 x - \left\{ {\sqrt 2 x} \right\}} \right) =   \cr 
  &  = \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\} \cr} 
} \tag {2}$$
and, since the fractional part ranges within $[0,1)$, the radicand is bound to 
$$ \bbox[lightyellow] {  
 - \,\sqrt 2  < r = \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\} < 1
} \tag {3}$$
If the function is defined in the real field, then its Domain of definition will be given by the values of $x$ such that
$$ \bbox[lightyellow] {  
x:\;\;0 \le \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\}\quad  \Leftrightarrow \quad \sqrt 2 \left\{ x \right\} \le \left\{ {\sqrt 2 x} \right\}
} \tag {4}$$
It can be easily seen that the fractional part of a sum can be written as
$$ \bbox[lightyellow] {  
\eqalign{
  & \left\{ {z + y} \right\} = \left\{ {\left\{ z \right\} + \left\{ y \right\}} \right\} =   \cr 
  &  = \left\{ z \right\} + \left\{ y \right\} - \left\lfloor {\left\{ z \right\} + \left\{ y \right\}} \right\rfloor  =   \cr 
  &  = \left\{ z \right\} + \left\{ y \right\} - \left[ {1 \le \left\{ z \right\} + \left\{ y \right\}} \right] \cr} 
} \tag {5}$$
so that the RHS of the inequality (4) above becomes
$$ \bbox[lightyellow] {  
\eqalign{
  & \left\{ {\sqrt 2 x} \right\} = \left\{ {\sqrt 2 \left\lfloor x \right\rfloor  + \sqrt 2 \left\{ x \right\}} \right\} =   \cr 
  &  = \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right] \cr} 
} \tag {6}$$
and  the inequality transforms into
$$ \bbox[lightyellow] {  
\eqalign{
  & \sqrt 2 \left\{ x \right\}\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right]  \cr 
  & \quad  \Downarrow   \cr 
  & \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor  + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right]  \cr 
  & \quad  \Downarrow   \cr 
  & 0 \le \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor  + \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right]\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} < 1 \cr} 
} \tag {7}$$
Now, the Poisson bracket cannot be one (because of the <1), so it shall be null, i.e. its inequality false.
Also, the floor before it (which is either $0$ or $1$) cannot be but $0$.   
Thus we can conclude that the Domain of definition (y real) is given by
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} < 1 \hfill \cr 
  \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor  = 0 \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  \left( {0 \le } \right)\left\{ x \right\} < 1/\sqrt 2  \hfill \cr 
  \left( {0 \le } \right)\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \sqrt 2 \left\{ x \right\} < 1 \hfill \cr}  \right.
} \tag {7.a}$$
and the Codomain is 
$$ \bbox[lightyellow] {  
0 \le y < 1
} \tag {7.b}$$
Because of the irrationality of $\sqrt{2}$ inequality (7.a) does not add much to the original (4), and cannot be further simplified. 
A: At first I tried some manipulation but could not complete it. If we take $x=n+\sqrt{p}$ where $0 \leq p<1$ we obtain
$$\sqrt{2}n-[\sqrt{2}n+\sqrt{2p}]$$
we can deduce that $p<0.5$. 
Now if we take $x=n+p$ and $\sqrt{2}=1+p_1$ (where $p_1$ is about $0.41$, also this can be considered for $\sqrt{y}[x]-[\sqrt{y}x]$ with $y=m+q$ where $0\leq q<1$) we have
$$(1+p_1)\times[n+p]\geq [(1+p1)\times(n+p)] $$
$$n+np_1\geq[n+np_1+p+pp_1]$$
$$np_1\geq[np_1+p(1+p_1)] \quad or \quad [p_1(n+p)+p]$$
from inequalities above 
$$[np_1+p(1+p_1)]=[np_1]$$
if we take $[np_1]=m$ and $r=np_1-m$ we can obtain
$$p(1+p_1)<1-r=1-np_1+[np_1]$$
So that
$$p<\frac{1-np_1+[np_1]}{1+p_1}$$
For example for $n=1$ we have:
$$0\leq p<\sqrt{2}-1$$
For $n=2$, $0\leq p<\frac{3-2\sqrt{2}}{\sqrt{2}}$ which is about $0.1213$. The domain is:
$$\bigcup {[n,n+p]},\quad p<\frac{1-np_1+[np_1]}{1+p_1}$$
I tried both with MATLAB and it is true. The domain depends on the number $n$ and also the decimals of $\sqrt{2}$, $p_1$ which are known. This can be generalized to domain of $\sqrt{\sqrt{y}[x]-[\sqrt{y}x]}$.
