Proof related to absolute value

Question

I was trying to prove

$|x||y| = |x\cdot y|$ but do not have a clue to start. I have seen examples of |x|+|y| >= |x+y| but could not translate it to my problem. Please my fellow math geniuses, help a brother out. Much appreciated

Answer 1

It is obvious that if either $x=0$ or $y=0$, then $xy=0$ and $|xy|=0$, but also $|x||y|=0$, so the statement holds. Now, what if both $x$ and $y$ are nonzero?

If $x$ and $y$ are real numbers, then, as David Peterson wrote in his comment, you only have to separate $4$ cases:

• $x>0,y>0$
• $x>0, y<0$
• $x<0, y>0$
• $x<0, y<0$

In each case, proving that $|x||y|=|xy|$ is simple.

If $x$ and $y$ are real numbers, then the easiest way is to write $x=r_1e^{i\phi_1}$ and $y=r_2e^{i\phi_2}$, however, this is not the most basic way. The basic way is to write $x=a+bi$ and $y=c+di$, then using the fact that $|a+bi|=\sqrt{a^2+b^2}$ to prove that $|xy|=|x||y|$.

Answer 2

If $x=y=0$, then

$$ |x|\cdot |y| = 0 \cdot 0 = 0 = |0| = |xy| $$

Suppose now $x >0 $ and $y < 0 $, ($xy < 0$) then

$$ |x| \cdot |y| = x \cdot (-y) = -xy = |xy|$$

Similarly, you can check the case $x < 0$ and $y > 0$. The case $x,y >0$ is proved in the comment to your question.