Jtdylktuy

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 $|a|= age 0$. 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 $aoverline ain mathbb R^+$ (see proof) then $|a| = sqrt{aoverline a}$.

Acceptence proof that $aoverline ain mathbb R^+$: First if $a = e + di$ where $e,d$ are real then $aoverline a =(e+di)(e-di) = e^2 +edi - edi - d^2i^2 = e^2 -(-d^2) = e^2+d^2$. $e^2$ and $d^2$ are both non-negative real numbers so $e^2 + d^2$ is too.

So $ain mathbb C$ 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 $kge 0$.)

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 $mge 0$ 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 $kge 0$ 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.

Go by the basic definition of the modulus function.

$|x|=begin{cases}x,&xge0\-x,&x<0end{cases}$

Since $|x|ge0, Big||x|Big|=|x|$

By definition, $|x| = x$ when $x ge 0$, and $|x| = -x$ when $x < 0$. Since $|a|$ is non-negative, $big||a|big| = |a|$.