Proving a property of modulus function by exhaustion: $\forall\ x \in\mathbb{ R} : |xy| = |x||y| $ I would like some clarification of a quick proof of the properties of the modulus function to make sure I'm doing the right thing.
$$\forall\ x \in\mathbb{R}  : |xy| = |x||y| $$
If I let $ x,y \in\mathbb{ R} $ and prove by exhaustion do I do this by considering when $x \ge 0, y\ge0$  then $|xy| = |x||y|$ and when $x < 0, y<0$  then $|xy| = |-x||-y| = |x||y|$.
Is this the correct way to go about it?
 A: That works, but then to complete the proof, you also need to consider, 


*

*the cases where $\;x \geq 0, \;y\leq 0,\;$ and by symmetry: $\;x \leq 0,\; y\geq 0$.


Since there are only four cases to consider, proof by exhaustion works just fine.
A: yes, but don't forget the case where one argument is positive and the other is negative. In proofs by exhaustion you must make sure you really did exhaust all cases. 
Just for the sake of pleasing the most pedantic among us, the proof will look nicer if you write it like this: Assume $x\ge 0$ and $y\ge 0$. Then $xy\ge 0$, and thus, by definition, $|x|=x$ and $|y|=y$ and $|xy|=xy$. So, $|xy|=xy=|x||y|$. 
Similarly for the other cases. 
A: In the interest of providing a complete proof (i.e. to expand Ittay Weiss' answer in its full, pedantic detail):
Note that
$$|x| := \begin{cases}
x & \text{if $x \ge 0$, and} \\
-x & \text{if $x < 0$,} \\
\end{cases}
\qquad\text{and}\qquad
|y| := \begin{cases}
y & \text{if $x \ge 0$, and} \\
-y & \text{if $x < 0$.} \\
\end{cases} $$
Hence there are four cases to consider: $x$ can be either positive or negative, and $y$ can be either positive or negative.  The cases are addressed as follows:
Terse Version


*

*$x\ge 0$ and $y \ge 0$.
$$ |x||y| = xy \ge 0 \implies xy = |xy|. $$

*$x \ge 0$ and $y < 0$.
If $x = 0$, then $|x||y| = 0 = |xy|$.  Otherwise, $x>0$, so
$$ |x||y| = x(-y) = -(xy) > 0 \implies xy < 0 \implies |xy| = -(xy). $$

*$x < 0$ and $y \ge 0$.  Case (2), mutatis mutandis.

*$x < 0$ and $y < 0$.
$$ |x||y| = (-x)(-y) = xy > 0 \implies |xy| = xy. $$
In More Detail


*

*Suppose that $x \ge 0$ and $y\ge 0$.  Then, by definition of the absolute value, $|x| = x$ and $|y|=y$.  Hence
$$ |x| |y| = xy. $$
Since both $x$ and $y$ are nonnegative, their product is nonnegative as well.  That is, $xy \ge 0$.  From the definition of the absolute value, $|xy| = xy$.  Combining this with the above gives
$$ |x| |y| = xy = |xy|, $$
which is the desired result.

*Suppose that $x \ge 0$ and $y < 0$.  There are actually two subcases to consider here:  either $x = 0$ or $x > 0$.  In the first case, if $x = 0$, then
$$ |x||y| = 0 = |xy|, $$
which is the desired result.  In the second case, $x > 0$.  Then
$$ |x||y| = x(-y) = -(xy) $$
(by the commutativity of multiplication; write $-y = (-1)y$). But $0 < |x||y| = -(xy)$, so $xy < 0$.  Hence
$$ |xy| = -(xy). $$
Combining this with the above gives
$$ |x||y| = -(xy) = |xy|, $$
which is again the desired result.

*Suppose that $x < 0$ and $y \ge 0$.  This case can be argued in exactly the same way as the previous case, replacing every instance of $x$ with $y$, and vice versa.

*Suppose that $x < 0$ and $y < 0$.  Then
$$ |x| |y| = (-x)(-y) = xy. $$
Since both $x$ and $y$ are negative, $xy > 0$.  Therefore $|xy| = xy$.  Hence
$$ |x||y| = xy = |xy|, $$
which is the desired result.

