<s>[INST] 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.
 [/INST] Well, by the trivial inequality, we have that $x^2$ is nonnegative. Note that you are solving for the floor of the $y$ outputs, not the $x$ inputs. Also, $\text{floor}(-x)^2=\text{floor}(x^2),$ so both sides are symmetrical.
</s>