Proof-Verification: $|xy|=|x|\cdot |y|$

Question

Proof-Verification: $|xy|=|x|\cdot |y|$

I've made a distinction in several cases. So for:

a) $x,y\geq 0 \Rightarrow x\cdot y\geq 0$:

$|x\cdot y|:=\begin{cases}x\cdot y , \quad x\cdot y \geq 0 \\ -(x\cdot y) , \quad x\cdot y <0\end{cases}$ So for $x\cdot y\geq 0$, we conclude $|x\cdot y|=x\cdot y$. Furthermore, we use the definitions for $|x|$ and $|y|$ to conclude $|x|\cdot |y|=x\cdot y|$.

b) $x,y<0 \Rightarrow x\cdot y \geq 0$:

Similarly, we have $|x\cdot y|=x\cdot y$ and by the definitions of $|x|$ and $|y|$, we have $|x|\cdot |y|=(-x)(-y)=x\cdot y$.

Hence the entire statement is proven. $\Box$

Is that a valid proof?

Answer 1

Here it is an alternative approach:

\begin{align*} |xy|^{2} = (xy)^{2} = x^{2}y^{2} = |x|^{2}|y|^{2} \Longleftrightarrow |xy| = |x||y| \end{align*}

Answer 2

Equality true for $x=0$ or $y=0$.

Let $x, y \not =0$.

$0=(xy)^2-(xy)^2=$

$ |xy|^2-x^2y^2=|xy|^2-|x|^2|y|^2=$

$(|xy|-|x||y|)(|xy|+|x||y|)$;

For $x,y \not =0$:

We have $|xy|=|x||y|$, since $(|xy|+|x||y|) >0$.

Used: $a^2=|a|^2.$