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  "body": "\n## Proof\n\nWe take the supremum of the numbers whose square is at most $x$, and rule out the two strict inequalities by contradiction.\n\n1. Let $x$ be a real number with $x \\ge 0$. All computations take place in the [[ordered field|0QSPNASE50]] $\\R$.\n2. Lemma. If $0 \\le a \u003c b$, then $a^2 \u003c b^2$. Indeed, $b^2 - a^2 = (b - a)(b + a)$, with $b - a \u003e 0$ and $b + a \\ge b \u003e 0$. A product of two positive numbers is positive.\n3. Uniqueness. Let $y, z \\ge 0$ with $y^2 = z^2 = x$. If $y \u003c z$, the lemma gives $y^2 \u003c z^2$, which is false. In the same way $z \u003c y$ is false. So $y = z$.\n4. Existence. Let $S = \\set{t \\in \\R : t \\ge 0 \\text{ and } t^2 \\le x}$. Then $0 \\in S$, so $S$ is nonempty.\n5. The set $S$ is [[bounded above|0B5H2V4JW8]] by $x + 1$. If $t \u003e x + 1 \\ge 0$, the lemma gives $t^2 \u003e (x + 1)^2 = x^2 + 2x + 1 \u003e x$, so $t \\notin S$.\n6. By the [[completeness axiom|0AXWET80YW]], $S$ has a [[supremum|0BTMQ04556]] $y$. Since $0 \\in S$, $y \\ge 0$.\n7. Suppose $y^2 \u003c x$. Let $\\delta = x - y^2 \u003e 0$ and $h = \\min(1, \\delta/(2y + 1))$. Then $0 \u003c h \\le 1$, so $h^2 \\le h$.\n8. Then $(y + h)^2 = y^2 + 2 y h + h^2 \\le y^2 + h (2y + 1) \\le y^2 + \\delta = x$.\n9. So $y + h \\in S$, and $y + h \u003e y$. This contradicts that $y$ is an upper bound of $S$.\n10. Suppose $y^2 \u003e x$. Then $y \\ne 0$, since $0^2 = 0 \\le x$. So $y \u003e 0$.\n11. Let $\\delta = y^2 - x \u003e 0$ and $h = \\min(y/2, \\delta/(2y))$. Then $h \u003e 0$ and $y - h \\ge y/2 \u003e 0$.\n12. Then $(y - h)^2 = y^2 - 2 y h + h^2 \\ge y^2 - 2 y h \\ge y^2 - \\delta = x$.\n13. Let $t \\in S$. If $t \u003e y - h$, then the lemma gives $(y - h)^2 \u003c t^2 \\le x$, against step 12. So $t \\le y - h$.\n14. Thus $y - h$ is an upper bound of $S$ smaller than $y$. This contradicts that $y$ is the least upper bound.\n15. By steps 9 and 14, neither $y^2 \u003c x$ nor $y^2 \u003e x$. Since the order is total, $y^2 = x$, and $y \\ge 0$.\n\n## Idea\n\nThe supremum $y$ of the numbers with square at most $x$ is the candidate. If $y^2$ were too small, a slightly larger number would still be in the set. If $y^2$ were too large, a slightly smaller number would still be an upper bound.\n",
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