How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$. 
How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.

I know that by definition $(\lfloor{x}\rfloor=m) ≡ (m≤x<m+1)$, so it follows that $(\lfloor{x^2}\rfloor=m ≡ (m≤x^2<m+1) ≡ (\sqrt{m}≤x^2<\sqrt{m+1})$; from this I can do 
$\lfloor{x^2}\rfloor=
\begin{cases}
...\\
-3, & -3≤x^2<-2 \\
-2, & -2≤x^2<-1 \\
-1, & -1≤x^2<0 \\
0, & 0≤x^2<1 \\
1, & 1≤x^2<2 \\
2, & 2≤x^2<3 \\
3, & 3≤x^2<4 \\
...
\end{cases}
$
=
$\begin{cases}
...\\
-3, & -3≤x^2<-2 \\
-2, & -2≤x^2<-1 \\
-1, & -1≤x^2<0 \\
0, & 0≤x<1 \\
1, & 1≤x<\sqrt{2} \\
2, & \sqrt{2}≤x<\sqrt{3} \\
3, & \sqrt{3}≤x<2 \\
...
\end{cases}
$
Everything is normal until I "solve" for x in the negative intervals of the domain, which results in this 
$\lfloor{x^2}\rfloor=
\begin{cases}
...\\
-3, & 3i≤x<2i \\
-2, & 2i≤x<i \\
-1, & i≤x^2<0 \\
0, & 0≤x<1 \\
1, & 1≤x<\sqrt{2} \\
2, & \sqrt{2}≤x<\sqrt{3} \\
3, & \sqrt{3}≤x<2 \\
...
\end{cases}
$
I don't know how to graph between those intervals.
Nonetheless when i see a graph of this function it seems that the function is symmetrical to the y axis. 

Could you explain me, please, how can I graph the function in those negative intervals?
Thanks in advance.

Guide:


*

*There are no real $x$ which satisfies $-1 \le x^2 < 0$.

*$0 \le x^2 < 1$ is equivalent to $0 \le |x|^2 < 1$ and the solution is $|x| < 1$, that is $-1 < x < 1$. 

*$1 \le x^2 < 2$ is equivalent to $1 \le |x|^2 < 2$ and the solution is $1 \le |x| < \sqrt{2}$, that is $\{ x: -\sqrt{2} < x \le -1\} \cup \{ x: 1 \le x < \sqrt2\}$. 
Note that the function is an even function.
