<s>[INST] 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 $y = \mid x \mid$?
 [/INST] It is the definition of square root of a number. The square root is defined in the sense that $s(x^2) = \sqrt{ x^2 } = |x|$ for all real $x$. Thus, the domain is the real numbers and the codomain is the non-negative real numbers. The reason it is defined this way is to make sure that $s$ is a function. Assume for a minute that $s(x^2) = x$, then:
$$\sqrt{(-5)^2} = -5, \qquad \sqrt{5^2} = 5$$
But we know that $\sqrt{(-5)^2} = \sqrt{25} = \sqrt{5^2}$. Thus we see that $s(25) = -5, 5$. And thus $s$ is not a function. To keep it as a function, we have to 'sacrifice' and say that $s(x^2) \neq x$. Rather, $s(x^2) = |x|$. This will be consistent with the definition of a function.
Since it is a definition, it cannot be proven. The problem is that many think that $\sqrt {x^2} = x$ because we study positive numbers before we study negative numbers, which is understandable, because I used to make that mistake all the time.
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