Square root of negative integer

Question

Can I write:

$-\sqrt{(2)}$ = $\sqrt{(-2)}$

and vice versa?

Or, say, we have,

$(-\sqrt{(x - 4)}$

Can this be changed into $(\sqrt{(4 - x)}$ by taking the minus sign inside the square root? How?

Answer 1

Consider the following logic if your first was correct:

$1 = (-1)(-1) = (-\sqrt1)(-\sqrt1)= \sqrt{-1}\sqrt{-1}=-1$ (As $\sqrt{x}\sqrt{x}=x$ under the normal definition)

For most people, such a contradiction is a bad bad thing.

Answer 2

We know that $ a = \sqrt{x}$ if and only if $a^2 = x$, and so if we have the square root of a negative number we are looking for an $a$ that if squared will equal a negative number. Now if $a \in \mathbb{R}$ we can have either a negative $a$ or a positive $a$. If a is negative we can square it to get $$ (-a)^2 = (-a)(-a) = (-1)(-1)(a)(a) = (a)(a) = a^2 $$ and so we see that no matter what, so long as $a \in \mathbb{R}$, we can only have positive squares, and therefore can only take square roots of positive numbers. Now if you do want to take the square root of a negative number you must find an $a$ that when squared equals a negative number. However, because we are looking for $a = \sqrt{-x}$ and we know that that $$\begin{align} a &= \sqrt{-x}\\ &= \sqrt{-1}\sqrt{x} \end{align}$$ we just need to find $\sqrt{-1}$ in order to find the square root of any $-x$. Here we run into a problem, because no $a \in \mathbb{R}$ can fit our description. We now define $\sqrt{-1} = i$, but $ i \not\in \mathbb{R}$. With this new element we can now form a new type of number, namely complex numbers (this is their name, not me calling them complicated), of the form $a + bi, a, b \in \mathbb{R}$ This is called doing a field extension, but that is just a technical term for creating more numbers outside the numbers you already have, in essence you are extending your number system. Now we can take square roots of negative numbers!