Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:
$\begin{align}
f(x) &= \sqrt {x^2} \\
f(-1) &= \sqrt {(-1)^2} \\
f(-1) &= \sqrt {1} \\
f(-1) &= 1
\end{align}$
$\begin{align}
g(x) &= x \\
g(-1) &= -1
\end{align}$
thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:
$y = \sqrt {x^2}$:

$y = x$:

$y = \mid x \mid$:

From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$?

When $x < 0$, $|x| = -x > 0$.  $-x \ne x$ (unless $x=0$) and $(-x)^2 = x^2$.   There are two "square roots" of any positive number $y$, i.e. numbers whose square is $y$, and the positive one is called $\sqrt{y}$. So $\sqrt{x^2} = -x = |x|$ when $x < 0$, and $\sqrt{x^2} = x = |x|$ when $x \ge 0$.
