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  "body": "\n## Proof\n\n1. Let $x$ and $y$ be real numbers with $x \u003c y$. Then $y - x \u003e 0$.\n2. By the [[Archimedean property|0WH916TJNY]], there is a natural number $n \\ge 1$ with $1/n \u003c y - x$.\n3. Let $k = \\floor{n x}$ be the [[floor|0YVRNEDSVG]] of $n x$, and let $m = k + 1$. Then $m$ is an integer.\n4. By the [[characterization of the floor|0R9B0B5458]], $k \\le n x \u003c k + 1$. That is, $m - 1 \\le n x \u003c m$.\n5. Divide by $n \u003e 0$, which keeps the order in the [[ordered field|0QSPNASE50]] $\\R$. This gives $m/n - 1/n \\le x \u003c m/n$.\n6. So $x \u003c m/n$. Also $m/n \\le x + 1/n \u003c x + (y - x) = y$, by steps 2 and 5.\n7. Let $q = m/n$. It is a [[rational number|0NCJ5WXN6Z]], since $m$ and $n$ are integers and $n \\ne 0$. By step 6, $x \u003c q \u003c y$.\n\n## Idea\n\nChoose a step size $1/n$ smaller than the gap $y - x$. The grid of fractions $m/n$ cannot jump over the gap, so the first grid point to the right of $x$ is still below $y$.\n",
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