How to prove that $\big| |a| \big| = |a|$? I mean it is somehow obvious as squaring makes numbers positive and the square root is defined as a positive number, but I would appreciate a (long) answer.
Probably by stating that $\sqrt{a^2} = \sqrt{\left(\sqrt{a^2}\right)^2}$.
Basically as $|m|$ is always positive then $||a||$ is the absolute value or a positive number. And as absolute values maintain magnitude, it is a positive number with the same magintude as a positive number. I.e. itself.
If you use a formal definition such as:
Definition 1: If $a \in \mathbb R$ and and $a \ge 0$ then $|a| = a$. If $a < 0$ then $|a|=-a$.
Immediate Proposition: $|a| \ge 0$ always.
Proof: If $a \ge 0$ then $|a|= a\ge 0$. If $a < 0$ then $|a| = -a > 0$.
So if $|a| \ge 0$ (which is always true) then $||a|| = |a|$. DOne.
Definition 2: If $a \in \mathbb C$ and if we accept that $a\overline a\in \mathbb R^+$ (see proof) then $|a| = \sqrt{a\overline a}$.
Acceptence proof that $a\overline a\in \mathbb R^+$: First if $a = e + di$ where $e,d$ are real then $a\overline a =(e+di)(e-di) = e^2 +edi - edi - d^2i^2 = e^2 -(-d^2) = e^2+d^2$. $e^2$ and $d^2$ are both non-negative real numbers so $e^2 + d^2$ is too.
So $a\in \mathbb C$ then $|a| = k$ for some non-negative real number. Then $|a| = k + 0i$ and $\overline {|a|} = k - 0i=k$ and $|a|\overline{|a|}= k*k=k^2$. And so $||a|| = \sqrt{k^2} = k=|a|$ (because $k\ge 0$.)
Now there are a lot of basic axioms and propositions I have taken for granted:
I assume without verification that: For any real $m$ that $m^2 \ge 0$; and that: for any real $m\ge 0$ there exists a unique non-negative real number that I call $\sqrt{m}$ so that $(\sqrt{m})^2= m$.
[It's trivial to verify that if $k\ge 0$ then $\sqrt{k^2} =k$. That follows as there is a unique number $\sqrt{k^2}$ so that $\sqrt{k^2}^2 = k^2$ and $k$ is such a number such that $k^2 = k^2$. Since such a number is unique $\sqrt{k^2} = k$.]
....
You ask in comments how to verify that a negative times a negative is negative:
1) $0*m = 0$. Pf: $0*m = (0+0)m = 0*m + 0*m$ so $0 = 0*m - 0*m = 0*m + 0*m - 0*m = 0*m$.
2) $-(ab)=(-a)b= a(-b)$. $0b = (a + (-a))b = ab + (-a)b$ as $(-a)b = -(ab)$. The rest is the same.
Axiom: If $a< b$ then $a+c < b+c$ for all $c$.
3) If $ a> 0$ then $-a < 0$ and if $a< 0$ then $-a> 0$. Pf: $a> 0$ then $a+(-a) > 0 +(-a) $ so $0 > -a$. Ther rest is the same.
Axiom: If $a < b$ and $c > 0$ then $ac < bc$.
4)positive times positive is positive: Pf: Let $a > 0; b> 0$. Then $ab > 0b =0$.
5) Positive times negative is negative. Pf: Let $a > 0$ and $b<0$ then $-b > 0$ so $a(-b)=-ab > 0$ and so $ab < 0$.
6) Negative times a negative is positive: Pf: Let $a < 0$ and $b< 0$ then $-a > 0$ and $-ab < 0$ and so $-(-ab) > 0$ and so $ab > 0$.....
Oops I forgot :
3 $\frac 12$) $-(-m) = m$. Pf: $m+ (-m) = 0$ so $m = -(-m)$.
The final thing I took for granted was that if $k \ge 0$ then there exists a unique $a \ge 0$ so that $a^2= k$. We call that $a$ the radical of $k$ or $\sqrt{k}$.
To prove that we need some definitions about the completenes of the real numbers and I'm not going to get into that heare.