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  "body": "\n## Proof\n\n1. Let $x$ be a real number. By definition, the [[floor|0YVRNEDSVG]] $\\floor{x}$ is the greatest [[integer|0C4JS9RF5M]] $k$ with $k \\le x$.\n2. Let $n = \\floor{x}$. Then $n \\le x$.\n3. The integer $n + 1$ is larger than $n$. Since $n$ is the greatest integer at most $x$, $n + 1 \\le x$ fails. The order of $\\R$ is [[total|0QSPNASE50]], so $x \u003c n + 1$.\n4. Conversely, let $n$ be an integer with $n \\le x \u003c n + 1$.\n5. Let $m$ be an integer with $m \\le x$. Then $m \\le x \u003c n + 1$, so $m - n \u003c 1$.\n6. The number $m - n$ is an integer less than $1$, so $m - n \\le 0$. Thus $m \\le n$.\n7. By steps 4 and 6, $n$ is an integer at most $x$ and is at least every such integer. So $n$ is the greatest one, that is, $n = \\floor{x}$.\n8. Steps 2, 3 and 4 to 7 prove the equivalence. Taking $n = \\floor{x}$ in steps 2 and 3 gives $\\floor{x} \\le x \u003c \\floor{x} + 1$.\n\n## Idea\n\nThe integer $n + 1$ is the next integer after $n$. So $n$ is the greatest integer below $x$ exactly when $n$ is at most $x$ and $n + 1$ is not.\n",
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