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  "body": "\n## Proof\n\n1. Let $F$ be an [[Archimedean ordered field|0HQARMKEV5]] and let $x \\in F$.\n2. First, $0 \\le 1$ in $F$. The order is total, so $0 \\le 1$ or $1 \\le 0$.\n3. If $1 \\le 0$, add $-1$ to both sides, using the first axiom of an [[ordered field|010JCV9J4E]]. This gives $0 \\le -1$.\n4. Then the second axiom gives $0 \\le (-1)(-1) = 1$. So $0 \\le 1$ in both cases.\n5. In a field $1 \\ne 0$. Hence $0 \u003c 1$.\n6. Apply the definition of an Archimedean field with $y = 1$. It gives a natural number $n$ with $n \\cdot 1 \u003e x$.\n7. Here $n \\cdot 1 = 1 + \\dots + 1$, with $n$ terms, is the element $n$ of $F$. So $x \u003c n$.\n\n## Idea\n\nThe definition says that copies of any positive $y$ add up past any $x$. Taking $y = 1$, which is positive in each ordered field, gives the claim.\n",
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