Spivak Chapter 1 Problem (i) Prove $$\left| xy \right| = \left| x \right|   \cdot \left| y \right|$$
Spivak's proof:

$(\left| xy \right|)^{2} = ( xy )^{2} =  x ^{2} y ^{2} = (\left |x\right | ^{2} \left | y\right | ^{2}) = (\left |x  \right | \cdot \left | y \right|)^{2}$


Since $\left| xy \right|$ and $\left |x  \right | \cdot \left | y \right|$ are both $\geq 0$, this proves that $\left| xy \right| = \left| x \right|   \cdot \left| y \right|$

This proof forms the basis for most of the proofs that follow, so I really want to carefully understand a few concepts.

*

*What allows us to assume that $\left|xy \right |$ is $\geq 0$? Of course the construct $\left|xy \right |$ tells us this is the absolute value of $xy$, and indeed the definition shows us that $$\left| xy \right| = \begin{cases}
 xy, \qquad xy \geq 0 \\ 
 -xy, \qquad xy \leq 0
\end{cases} $$ but that is not what Spivak says. Can someone weigh in?


*I can more readily accept that $\left |x  \right | \cdot \left | y \right|$  is $\geq 0$ as this occurs subsequent to $(xy)^{2}$
This might seem rather trivial to most of the seasoned on this board, but getting a good foundation I think is crucial.  Thanks for your help, as always.
 A: As noticed in the comments $|xy|\ge 0$ and $|x|,|y|\ge 0$ can be easily proved separetely by the definition.
As an alternative proof we can show that

*

*for $xy=0 \implies 0=|xy|=|x||y|=0$

*for $xy<0 \implies -xy=|xy|=|x||y|=-x \cdot y=-xy\quad \lor \quad  -xy=|xy|=|x||y|=x \cdot -y=-xy$

*for $xy>0 \implies xy=|xy|=|x||y|=x \cdot y=xy \quad \lor \quad  xy=|xy|=|x||y|=-x \cdot -y=xy$
therefore $|xy|$ is equivalent to $|x||y|$.
A: From Calculus by Spivak, Chapter 1, Page 11 (3rd edition)

For any number $a$, we define the absolute value $|a|$ of $a$ as follows:
$$|a| =
\begin{cases}
a, & a \ge 0 \\
-a, & a \le 0.
\end{cases}$$
Note that $|a|$ is always positive, except when $a = 0$.

This holds for any real number, including $x$, $y$, and $xy$.
$|xy| \ge 0$ by definition.
As for your comment that

I can more readily accept that $|x|\cdot|y|$ is $\ge 0$ as this occurs subsequent to $(xy)^2$

there is a different and I think simpler argument that you may be missing:
By definition $|x|$ and $|y|$ are each $\ge 0$. The product of two nonnegative numbers is always nonnegative because:

*

*If either one is zero, the product is zero ($0\cdot a = 0$ for any number $a$.)


*If neither is zero, they are both positive and so their product is positive (closure under multiplication).
