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  "body": "\n## Proof\n\n1. Let $a$ and $b$ be real numbers. For each real $t$, $\\abs{-t} = \\abs{t}$, by the definition of the [[absolute value|00S7C68D51]].\n2. By the [[triangle inequality|03HF5PMY5T]], $\\abs{a} = \\abs{(a - b) + b} \\le \\abs{a - b} + \\abs{b}$. So $\\abs{a} - \\abs{b} \\le \\abs{a - b}$.\n3. Exchange $a$ and $b$: $\\abs{b} - \\abs{a} \\le \\abs{b - a} = \\abs{a - b}$, by step 1.\n4. Let $s = \\abs{a} - \\abs{b}$. By steps 2 and 3, $s \\le \\abs{a - b}$ and $-s \\le \\abs{a - b}$.\n5. By definition, $\\abs{s}$ equals $s$ or $-s$. In both cases, $\\abs{s} \\le \\abs{a - b}$ by step 4.\n6. So $\\bigl| \\abs{a} - \\abs{b} \\bigr| \\le \\abs{a - b}$.\n\n## Idea\n\nWrite $a$ as $(a - b) + b$ and apply the triangle inequality. This bounds $\\abs{a} - \\abs{b}$, and the same argument with $a$ and $b$ exchanged bounds its negative.\n",
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