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  "body": "\n## Proof\n\nLet $\\mathcal{C}$ be the set of the [[constructible numbers|0FMN2MN7TM]]. We argue by contradiction.\n\n1. By the [[field theorem|0PN9WMRWT7]], $\\mathcal{C}$ is a subfield of $\\C$ closed under square roots.\n2. Suppose that $\\sqrt{\\pi} \\in \\mathcal{C}$. Then $\\pi = \\sqrt{\\pi} \\cdot \\sqrt{\\pi} \\in \\mathcal{C}$, since $\\mathcal{C}$ is closed under products.\n3. Every constructible number is [[algebraic over $\\Q$|0SSSM1D475]]. So $\\pi$ is algebraic over $\\Q$.\n4. This contradicts the [[transcendence of $\\pi$|0G0F7FGCGP]]. So $\\sqrt{\\pi} \\notin \\mathcal{C}$.\n\nNow we turn to the construction problem. Take the given circle with centre $0$ through $1$, so its area is $\\pi$. A [[squaring of the circle|0BSXRER81B]] would construct two points $P, Q$ with $\\abs{P - Q} = \\sqrt{\\pi}$, the side of a square of area $\\pi$.\n\n5. Let $\\overline{\\mathcal{C}}$ be the set of the [[conjugates|0BJV24C4SY]] of the elements of $\\mathcal{C}$. Conjugation preserves the field operations, so $\\overline{\\mathcal{C}}$ is a subfield of $\\C$.\n6. If $w = \\bar z$ with $z \\in \\mathcal{C}$ and $t^2 = w$, then $\\bar t^2 = z$, so $\\bar t \\in \\mathcal{C}$ and $t \\in \\overline{\\mathcal{C}}$. So $\\overline{\\mathcal{C}}$ is closed under square roots.\n7. The field theorem says that $\\mathcal{C}$ is the smallest subfield of $\\C$ closed under square roots. So $\\mathcal{C} \\subseteq \\overline{\\mathcal{C}}$. Each $z \\in \\mathcal{C}$ is then $\\bar u$ for some $u \\in \\mathcal{C}$, and $\\bar z = u \\in \\mathcal{C}$.\n8. Let $w = P - Q \\in \\mathcal{C}$. Then $\\bar w \\in \\mathcal{C}$, so [[$w \\bar w = \\abs{w}^2$|0ZB7J071WQ]] is in $\\mathcal{C}$.\n9. Its square root, the [[modulus|0KW2X0TFBX]] $\\abs{w} = \\sqrt{\\pi}$, is in $\\mathcal{C}$. This contradicts step 4.\n10. So $\\sqrt{\\pi}$ is not constructible, and the circle cannot be squared with straightedge and compass.\n\n## Idea\n\nConstructible numbers are algebraic, and they form a field. If $\\sqrt{\\pi}$ were constructible, its square $\\pi$ would be algebraic, which Lindemann's theorem rules out. A squaring of the circle would make the length $\\sqrt{\\pi}$ constructible.\n",
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