Problematic square root

Question

Ok, here is what I think. Please correct me if I am wrong. $$\sqrt{9} \neq 3$$ and also $$\sqrt{9} \neq -3$$

Now let's assume, that above statements are false, then we have $-3 = \sqrt{9} = 3$ and since $3 \neq -3$ the assumption must be wrong. Ok, square root must be equal to 3 and -3 at the same time. As of my understanding a set of numbers is not a number itself and that leads to a conclusion, that a square root of a number is not a number. Right?

Both answers are of the same importance - neither of them is superior. Then here is the question - when we plot $f(x) = \sqrt{x}$, why do we always plot positive answers? Can someone give me a proof, that plotting negative answers is not allowed?

Answer 1

It's not a mistake to plot the negative branch of the graph $\sqrt{x}$. Rather it arises from the convention that the sign $\sqrt{x}$ means the positive root of x, whilst $-\sqrt{x}$ means the negative root of x. If we wish our graph to represent a function, with the important property that it be single-valued (one value of $x$ gives a unique value of $y$), then we have to restrict the graph to one or the other of the branches.

Answer 2

You are partially correct:

Now let's assume, that above statements are false, then we have $−3=\sqrt{9}=3$ and since $3≠−3$ the assumtion must be wrong.

Yes. It is all correct. The key is that the prhase "above statements are false" means $\sqrt{9}\neq3$ AND $\sqrt{9}\neq-3$ are both false. Then, since this statement is false, the true is $\sqrt{9}=3$ OR $\sqrt{9}=-3$, which indeed is true cause the first is true ($\sqrt{9}$ means, by definition, the positive root of 9)

Answer 3

The equation $x^2 = 9$ does indeed have two solutions, and therefore it is not incorrect to say that "$9$ has two square roots", namely $x=3$ and $x=-3$. But the notation $\sqrt{9}$, by definition, refers to the positive square root of $9$. There is only one of those, and it is equal to $3$, not to $-3$.