Square roots: $ -\sqrt{a} = \sqrt{b}$

Question

[CLOSED] Thanks GoodDeeds and Henry [1] I understood the fundamental problem.

As √9 = ± 3

If , √A = √B
Thus, ** ±a = ±b**
And so a = b; -a = b; and a = -b;;
Thus, √9 = +3 OR -3

Let, √A = ±a
and, √B = ±b
Is this reasoning correct?

If not how does this actually work?

[1] \bib{26369}{misc}{
title={Square roots -- positive and negative},
author={Henry (https://math.stackexchange.com/users/6460/henry)},
note={URL: https://math.stackexchange.com/q/26369 (version: 2011-03-11)},
eprint={https://math.stackexchange.com/q/26369},
organization={Mathematics Stack Exchange}
}

Answer 1

No. The mistake is that $\sqrt{9}\ne\pm3$, $\sqrt{9}=+3$ only.

This is because the square root function is defined such that only the non negative root is taken. In general, $$\sqrt{x^2}=|x|$$

Moreover, your reasoning will lead to absurd conclusions, such as $$\sqrt{9}=\sqrt{9}$$ $$+3=-3$$ which is obviously not true.

Answer 2

As √9 = ± 3

No, the square root function is always $\ge 0$.

Thus, √9 = +3 OR -3

No, the square root function is always $\ge 0$.

Let, √A = ±a and, √B = ±b

No, the square root function is always $\ge 0$.

If , √A = √B Thus, ** ±a = ±b** And so a = b; -a = b; and a = -b;;

Is this reasoning correct?

The reasoning is not correct because the square root function is always $\ge 0$.

If not how does this actually work?

The square root function is always $\ge 0$.

Answer 3

The definition of the square root of $x \in \mathbb{R}, x \geq 0$ is the only positive number $y \in \mathbb{R}, y \geq 0$ such taht $y^2=x$.

Your reasoning works with a different definition of the square root, which is a function $yoursqrt : \mathbb{R} \rightarrow \mathbb{R}^2 $