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  "body": "\n## Proof\n\n1. Let $\\varepsilon$ be a real number with $\\varepsilon \u003e 0$.\n2. Since the [[reals are Archimedean|0A3GWZ86AY]], applied with $x = 1$ and $y = \\varepsilon$, there is a natural number $n$ with $n \\varepsilon \u003e 1$.\n3. We have $n \\ge 1$. Indeed, for $n = 0$ we would get $0 \\cdot \\varepsilon = 0 \u003e 1$, but $0 \u003c 1$ in each [[ordered field|010JCV9J4E]].\n4. In the [[ordered field|0QSPNASE50]] $\\R$, if $a \u003c b$ and $c \u003e 0$, then $a c \u003c b c$. Indeed, $(b - a) c \\ge 0$ by the product axiom, and $(b - a) c \\ne 0$ because both factors are nonzero.\n5. Since $0 \u003c 1$ and $n \\ge 1$, adding $1$ repeatedly gives $0 \u003c n$. The number $1/n$ is positive. It is nonzero. If $1/n \u003c 0$, then $-1/n \u003e 0$, and step 4 with $0 \u003c n$ gives $0 \u003c n \\cdot (-1/n) = -1$, which is false.\n6. Multiply $1 \u003c n \\varepsilon$ by $1/n \u003e 0$, using step 4. This gives $1/n \u003c \\varepsilon$.\n\n## Idea\n\nThe Archimedean property says that enough copies of $\\varepsilon$ exceed $1$. Dividing $n \\varepsilon \u003e 1$ by $n$ gives $1/n \u003c \\varepsilon$.\n",
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