Is $\sqrt{x^2} = x$? Does the $\sqrt{x^2}$ always equal $x$? I am trying to prove that $i^2 = -1$, but to do that I need to know that $\sqrt{(-1)^2} = -1$. If that is true then all real numbers are imaginary, because an imaginary number is any number that can be written in terms of $i$. For example, 2 can be written as $i^2 + 3$. Does this work or did I make an error?
 A: Not always. $\sqrt{(-1)^2}=\sqrt{1}=1\neq -1$. In general $\sqrt{x^2}=|x|$
A: *

*It is not true that $\sqrt{x^2} = x$.  As a very simple example, with $x=-2$, we obtain
$$ \sqrt{(-2)^2} = \sqrt{4} = 2 \ne -2. $$
In general, if $x \in \mathbb{R}$, then $\sqrt{x^2} = |x|$.  Things get more complicated when you start working with complex numbers, but I think that a discussion of "branches of the square root function" is quite a bit beyond the scope of this question.


*There is a serious problem of definitions in the question.  The question asserts "...all real numbers are imaginary, because an imaginary number is any number that can be written in terms of $i$."  However, this is not the definition of an imaginary number.  An imaginary number is a number $z$ such that there is some real number $y$ such that $z = iy$, where $i$ is the imaginary unit.  A number such as $i^2 + 3$ is not an imaginary number, since there is no real number $y$ such that $2 = i^2 + 3 = iy$.
On the other hand, it is reasonable to say that every real number is a complex number.  A complex number is a number $z$ such that there are $x,y\in\mathbb{R}$ such that $z = x + iy$.  In the case of the example give, we have
$$ i^2 + 3 = (-1) + 3 = 2 = 2 + i0. $$
