when $\sqrt{y}$ and $\sqrt{x}$ are defined, is $\sqrt{y}$ = $\sqrt{x}$ a function? when $\sqrt{y}$ and $\sqrt{x}$ are defined, is $\sqrt{y}$ = $\sqrt{x}$ a function? for (x,y) in the reals. I think I'm thinking to hard about what the graph will look like
 A: Not so hard to find. The graph is the half-line $y=x, x\geq 0, y\geq 0$. 
A: $\sqrt{y}=\sqrt{x}$ is not what you call a function, it is an equation.
The solution $y(x)$ to that equation over positive numbers $x\in\mathbb{R}^+$ is a function, it is 
\begin{align*}
y : &&\mathbb{R}^+ &\mapsto \mathbb{R}\\
&&x &\rightarrow x
\end{align*}
also called the identity function, here resticted to positive numbers. You easily see that  it is such that $\sqrt{y(x)}=\sqrt{x}$ for all $x$ in its domain $\mathbb{R}^+$, so this function solves the equation $\sqrt{y}=\sqrt{x}$.
You often abbreviate it as $y(x)=x$ but this is a shortcut for the above (and a shortcut that doesn't mention on which interval it operates, which can be a problem sometimes).
The solution to $y=\sqrt{x}$ over $\mathbb{R}^+$ would have been
\begin{align*}
y : &&\mathbb{R}^+ &\mapsto \mathbb{R}\\
&&x &\rightarrow \sqrt{x}
\end{align*}
and the one to $\sqrt{y}=x$ over $\mathbb{R}^+$ would have been
\begin{align*}
y : &&\mathbb{R}^+ &\mapsto \mathbb{R}\\
&&x &\rightarrow x^2
\end{align*}
A: Attempt:
1) $x,y \gt 0.$
We have $√y-√x=0:$
Recall: $y-x=(√y-√x)(√y+√x).$
Then:
$y-x=0$ $\iff$ $(√y-√x)=0$, since $√y+√x >0$.
2) $x=y=0$ , then trivially $√y=√x$, and $y=x$.
Hence for $x,y \ge 0$ we get the more familiar form of a function $y=x$.
