Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$? Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ ?
So why is it that my calculator said otherwise? And I know that you can't square root a negative number, because it will gave an imaginary number. Is it because the calculator assume that if it there is an undefined value in the equation, the final result will be undefined too? Or is there a proof that I aren't aware of?
Unnecessary comment below:
And if graph of $y=1$ aren't the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ just like the calculator said. That would be great, because you don't have to clarify the domain outside the expression. You could just do $y=2x\cdot\frac{\sqrt{4-x}}{\sqrt{4-x}}$ instead of using a set notation such as $\{x\in| \mathbb{R},x>4\}$
 A: If you define $y=1$ on all real numbers, including $4$, they cannot be the same as you cannot define the other function on the same domain, and for two functions to be the same (formally) their domain has to be the same.
On $\mathbb{C}$ the same holds, as for $x = 4$, the second function is not defined.
If you write these functions down formally you would have to say what there domain of definition is. If you define the first function $y=1$ on the same domain as the second they are the same. So it all depends on your choice of domain, which shows how important it is to keep in mind, that when defining a function you always have to state from where to where it maps, otherwise the definition is uncomplete.
I hope this also clarifies the question in your comment.
A: The equation $y=1$ is independent of $x,$ but your quotient of square roots is not. Working within the real numbers, the radical quotient makes sense only when $4-x > 0$ for two reasons: so you do not introduce complex numbers (if $4-x<0$); and so that the denominator is not $0$ (if $x=4$).
A: No
Your logic is sound in thinking that since $\frac{\sqrt{4-x}}{\sqrt{4-x}} = 1$, that $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ could be equal  to $y=1$, but since your value of $x$ isn't defined, in this case, $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ is not equal to $y=1$
