On the internet there are many proofs but very summary
Definition
$|a| = $
$a$ if $a \ge 0$
$-a$ if $a < 0$
Proposition 1
$|a| = 0 \Leftarrow \Rightarrow a = 0$
$\Rightarrow$)
Suppose that $|a| = 0$
This implies that $a\ge 0$ xor $a<0$
Case $a<0$
$|a| = -a$ by hypothesis $-a= 0$
As $a<0$
$a + (-a)<0 + 0$
$0<0$ (!contradiction)
Case $a \ge 0$
$|a| = a$ by hypothesis $a = 0$
$\Leftarrow$)
Suppose that $a= 0$
$|a| = a$ by hypothesis $|a| = 0$
proposition 2
$|a| \ge 0$
case $a \ge 0$
$|a| = a$ by hypothesis $|a| \ge 0$
case $a < 0$
$|a| = -a$
As $a<0$
$a+(-a)<0+(-a)$
$0<-a$
$0<|a| \Rightarrow 0 \le |a|$
are valid my proofs?