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  "body": "\n## Proof\n\nWe argue by cases on the signs of $a$ and $b$.\n\n1. Let $a$ and $b$ be real numbers. By the definition of the [[absolute value|00S7C68D51]], $\\abs{z} = z$ when $z \\ge 0$.\n2. Also $\\abs{z} = -z$ when $z \\le 0$. For $z \u003c 0$ this is the definition, and for $z = 0$ both sides are $0$.\n3. In the [[ordered field|0QSPNASE50]] $\\R$, the product of two nonnegative numbers is nonnegative. If $u \u003c 0$, then $-u \u003e 0$, by adding $-u$ to both sides.\n4. Case $a \\ge 0$ and $b \\ge 0$. Then $a b \\ge 0$, so $\\abs{a b} = a b = \\abs{a} \\abs{b}$.\n5. Case $a \\ge 0$ and $b \u003c 0$. Then $-b \u003e 0$ and $a (-b) \\ge 0$, so $a b \\le 0$. By step 2, $\\abs{a b} = -(a b) = a (-b) = \\abs{a} \\abs{b}$.\n6. Case $a \u003c 0$ and $b \\ge 0$. This is step 5 with $a$ and $b$ exchanged, since $a b = b a$.\n7. Case $a \u003c 0$ and $b \u003c 0$. Then $a b = (-a)(-b) \\ge 0$, so $\\abs{a b} = (-a)(-b) = \\abs{a} \\abs{b}$.\n8. The four cases cover all pairs $(a, b)$, since the order is total. So $\\abs{a b} = \\abs{a} \\abs{b}$.\n\n## Idea\n\nThe absolute value removes the sign. The sign of a product follows from the signs of the factors, and the sizes simply multiply.\n",
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