Absolute value proof

Question

Prove that $|xy|=|x||y|$

hint: consider the 4 cases, x and y both positive, both negative, x is positive while y is negative, and y is positive while x is negative.

I am not really sure how to go about proving this, especially the last 2 cases. A push in the right direction would be appreciated.

Proof: There are 4 cases.

Suppose that x>0 and y>0, then $|xy|=xy=|x||y|$

Suppose x<0 and y<0, then $|xy|=-x*-y=xy=|x||y|$

Suppose x<0 and y>0, then

Suppose x>0 and y<0 then

Answer 1

Case 3: Suppose that $x < 0$ and $y \geq 0$. Then $xy \leq 0$ so that: $$ |xy| = -xy = (-x)(y) = |x||y| $$ as desired.

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The other case is similar.

Answer 2

Hint : |$x$|$ = x$ for $x\ge 0$ $\ \ \ \ \ \ $ |$x$|$ =-x$ for $x<0$

Answer 3

For my taste, you're missing some connecting reasoning in the two cases you've already done. Let me show you the "both negative case" with a more solid argument:

Suppose $x$ and $y$ are both negative. Then $|x|=-x$ and $|y|=-y$. Also $xy$ is positive so $|xy|=xy$. Therefore, $$ |xy|=xy=(-x)\cdot(-y)=|x|\cdot|y| $$

The same scheme can be used for the other cases.