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  "body": "\n## Proof\n\n1. Let $t$ be real. We show $-\\abs{t} \\le t \\le \\abs{t}$, with the [[absolute value|00S7C68D51]].\n2. If $t \\ge 0$, then $\\abs{t} = t$ and $-\\abs{t} = -t \\le 0 \\le t$. If $t \u003c 0$, then $\\abs{t} = -t \u003e 0 \u003e t$ and $-\\abs{t} = t$.\n3. Let $r \\ge 0$ and let $s$ be real with $-r \\le s \\le r$. Then $\\abs{s} \\le r$. If $s \\ge 0$, then $\\abs{s} = s \\le r$. If $s \u003c 0$, then $\\abs{s} = -s \\le r$.\n4. In the [[ordered field|0QSPNASE50]] $\\R$, if $u \\le v$ and $u' \\le v'$, then $u + u' \\le v + u' \\le v + v'$, by adding $u'$ and then $v$.\n5. Let $a$ and $b$ be real. By step 2, $-\\abs{a} \\le a \\le \\abs{a}$ and $-\\abs{b} \\le b \\le \\abs{b}$.\n6. By step 4, $-(\\abs{a} + \\abs{b}) \\le a + b \\le \\abs{a} + \\abs{b}$.\n7. Here $r = \\abs{a} + \\abs{b} \\ge 0$. By step 3 with $s = a + b$, $\\abs{a + b} \\le \\abs{a} + \\abs{b}$.\n\n## Idea\n\nEach number lies between minus its absolute value and its absolute value. Adding two such bounds traps $a + b$ between $-(\\abs{a} + \\abs{b})$ and $\\abs{a} + \\abs{b}$.\n",
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