How can we prove that $\sqrt{ x^{2} }$ is equals to $|x|$?

Question

I used to use this equality at school. But now in my books of Analysis this property is not mentioned. Is this maybe incorrect?

Answer 1

In $\Bbb{R}$, the only numbers whose square is $x^2$ are $x$ and $-x$. Moreover, a square root is always positive. Since for $x\ge0,x=|x|$ and for $x\le0,|x|=-x$, we get the desired result.

Answer 2

All you need to show is that for all (real!) numbers $x$, the number $|x|$ is $\ge 0$ and its square equals $x^2$.

Answer 3

Why is a square root always non-negative?

Here is the reason.

$f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^2$ is neither one-one nor onto. However, if $\mathbb{R}_+$ represents the set of all non-negative real numbers, then $f:\mathbb{R_+}\to\mathbb{R_+}$ defined by $f(x)=x^2$ is both one-one and onto and consequently, it is bijective and invertaible.

Its inverse $f^{-1}:\mathbb{R_+}\to\mathbb{R_+}$ is defined by $f^{-1}(y)=\sqrt{y}$.

Thus, since the negative numbers are neither in the domain nor in the codomain of the square root function, the square root of a number can not be negative. In particular, $\sqrt{x^2}=|x|.$