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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $x$ and $y$ be real numbers with $y \u003e 0$. For a natural number $n$, write $n y = y + \\dots + y$ with $n$ terms. Then $(n + 1) y = n y + y$.\n2. Suppose that $n y \\le x$ for every natural number $n$.\n3. Let $S = \\set{n y : n \\in \\N}$. It is nonempty, and by step 2 it is [[bounded above|0B5H2V4JW8]] by $x$.\n4. By the [[completeness axiom|0AXWET80YW]], $S$ has a [[supremum|0BTMQ04556]] $b$.\n5. Since $y \u003e 0$, we have $b - y \u003c b$, in the [[ordered field|0QSPNASE50]] $\\R$.\n6. By the [[approximation property of the supremum|0M8TGGP9JP]], there is an element $m y$ of $S$ with $b - y \u003c m y$.\n7. Add $y$ to both sides: $b \u003c m y + y = (m + 1) y$.\n8. But $(m + 1) y \\in S$, so $(m + 1) y \\le b$ since $b$ is an upper bound. This contradicts step 7.\n9. So some natural number $n$ satisfies $n y \u003e x$. Thus $\\R$ is an [[Archimedean ordered field|0HQARMKEV5]].\n\n## Idea\n\nIf the multiples of $y$ were bounded, they would have a least upper bound $b$. Some multiple is within $y$ of $b$, and the next multiple then passes $b$, which is impossible.\n",
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