how to prove formally that the $\sqrt{x^2}$ is $|x|$, $x$ being a real number?

Question

Can the equality be proved like this?

(1) in case $x$ is positive (or equal to $0$), $\sqrt{x^2}$ is $x$.

(2) in case $x$ is negative, $\sqrt{x^2}$ is the positive number $-x$.

This tends to show that "$\sqrt{x^2}$" obeys the same conditions as $|x|$.

But I think it only rephrases what has to be proved, and does not really provide a justification.

In particular, how to explain (2)?

Answer 1

If $x$ is real, then $|x|$ is either $x$ or $-x$. Note that $|x|$ is always nonnegative.

• If $|x|$ is $x$, then $|x|^2=x^2$.
• If $|x|$ is $-x$, then $|x|^2=(-x)^2=(-1)^2x^2=x^2.$

So in any event, $|x|^2=x^2$.

Now $\sqrt{u}$ is defined to be the nonnegative number whose square is $u$. In particular, $\sqrt{x^2}$ is the nonnegative number whose square is $x^2$. We have just shown that this is $|x|$.

Answer 2

Let $t = x^2\ge 0$. Then $x=\pm\sqrt t$ and

$$|x| = |\pm \sqrt t| = \sqrt t = \sqrt {x^2}$$