Proof that the absolute value is multiplicative

Proof

We argue by cases on the signs of $a$ and $b$.

1. Let $a$ and $b$ be real numbers. By the definition of the absolute value, $\abs{z} = z$ when $z \ge 0$.
2. Also $\abs{z} = -z$ when $z \le 0$. For $z < 0$ this is the definition, and for $z = 0$ both sides are $0$.
3. In the ordered field $\R$, the product of two nonnegative numbers is nonnegative. If $u < 0$, then $-u > 0$, by adding $-u$ to both sides.
4. Case $a \ge 0$ and $b \ge 0$. Then $a b \ge 0$, so $\abs{a b} = a b = \abs{a} \abs{b}$.
5. Case $a \ge 0$ and $b < 0$. Then $-b > 0$ and $a (-b) \ge 0$, so $a b \le 0$. By step 2, $\abs{a b} = -(a b) = a (-b) = \abs{a} \abs{b}$.
6. Case $a < 0$ and $b \ge 0$. This is step 5 with $a$ and $b$ exchanged, since $a b = b a$.
7. Case $a < 0$ and $b < 0$. Then $a b = (-a)(-b) \ge 0$, so $\abs{a b} = (-a)(-b) = \abs{a} \abs{b}$.
8. The four cases cover all pairs $(a, b)$, since the order is total. So $\abs{a b} = \abs{a} \abs{b}$.