Is a product of two modulus functions the modulus of the product of the functions? I would just like to ask a simple question:
Is the following statement true? And is there a simple proof for it? $$|f(x)| \cdot|g(x)| = |f(x)\cdot g(x)| $$
 A: For $a\in\Bbb C$, by definition $\lvert a\rvert=\sqrt{a\overline a}$. Since, for non-negative real numbers $a,b$, the identity $\sqrt{ab}=\sqrt a\sqrt b$ holds, it can be observed that $$\lvert ab\rvert=\sqrt{ab\overline{ab}}\stackrel{\left(\overline{z\cdot w}=\overline w\cdot\overline z\right)}=\sqrt{ab\overline b\overline a}\stackrel{\left(w\overline w\in[0,\infty)\right)}=\sqrt{a\overline a}\sqrt{b\overline b}=\lvert a\rvert\lvert b\rvert$$
The special case where $a,b\in\Bbb R$ has the same passages with $a=\overline a$ and $b=\overline b$. A subtlety might be the fact that, for most textbooks, $\lvert a\rvert=\sqrt{a^2}$ is considered a theorem, rather than the definition of $\lvert a\rvert$. There is very good reason to make that choice, but it is in principle just a matter of taste.
A: Yes, it is true.
By definition (in $\mathbb R$) $$|x| =\begin{cases}x, &x\geq 0\\ -x, &x<0\end{cases} $$
Strictly speaking, there are four cases to consider, but effectively, we can make do by checking three cases.
1
Let $a,b\geq 0$, then $$ |ab| =ab = |a||b| $$
2
Let $a\geq 0, b<0$, then $$|ab| = -ab = a(-b) = |a||b| $$
For the last case, you would have to check this for $a<0$ and $b<0$.
A: Yes, $|a|\cdot|b|=|a\cdot b|$ holds for all real numbers. One proof is by considering the four cases 


*

*$a\ge 0, \, b\ge 0,$

*$a<0, \, b\ge 0,$

*$a\ge 0, \, b< 0,$

*$a<0, \, b< 0.$


It also holds for complex numbers, btw.
A: Use the fact that $|x| = x\cdot\text{sign}(x)$. Then,
$$|x||y| = \text{sign}(x)\text{sign}(y)(xy) = \text{sign}(xy)(xy).$$
