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  "body": "\n## Proof\n\nLet $\\mathcal{C}$ be the set of the [[constructible numbers|0FMN2MN7TM]]. A [[straightedge and compass|0DAECMZVRT]] step draws the line through two distinct points already obtained, or the circle centred at one obtained point through another. It then adds a common point of two distinct drawn lines or circles. Points constructed from points of $\\mathcal{C}$ lie in $\\mathcal{C}$, since the two sequences of steps can be joined.\n\nPart 1: basic constructions.\n\n1. Let $a \\in \\mathcal{C}$ with $a \\neq 0$. The line through $0$ and $a$ is $\\set{ta : t \\in \\R}$. The circle centred $0$ through $a$ meets it at $-a$, and the circle centred $a$ through $0$ meets it at $2a$. So $-a, 2a \\in \\mathcal{C}$.\n2. Let $a, b \\in \\mathcal{C}$ with $a \\neq b$, and let $m = (a+b)/2$. The points $u, v = m \\pm \\tfrac{\\sqrt 3}{2} i (b - a)$ satisfy $\\abs{u - a} = \\abs{u - b} = \\abs{b - a}$, and the same holds for $v$.\n3. So $u$ and $v$ lie on the circle centred $a$ through $b$ and on the circle centred $b$ through $a$. Hence $u, v \\in \\mathcal{C}$.\n4. The line through $u$ and $v$ is $\\set{m + t i (b - a) : t \\in \\R}$. It meets the line through $a$ and $b$ only at $m$. So the midpoint $m$ lies in $\\mathcal{C}$.\n5. Let $a, b \\in \\mathcal{C}$. If $a = b$, then $a + b = 2a \\in \\mathcal{C}$ by step 1, or $a + b = 0$. If $a \\neq b$, then $a + b = 2m$ with $m \\in \\mathcal{C}$ by step 4, so $a + b \\in \\mathcal{C}$.\n6. By steps 1 and 5, $\\mathcal{C}$ contains $0$ and $1$ and is closed under addition and negation.\n7. The circle centred $0$ through $1$ gives $-1 \\in \\mathcal{C}$. For $a = -1$ and $b = 1$, step 2 gives $u, v = \\pm \\sqrt 3 \\, i$, so the imaginary axis can be drawn. It meets the circle centred $0$ through $1$ at $i$, so $i \\in \\mathcal{C}$.\n8. Let $z \\in \\mathcal{C}$ not be real. Its [[conjugate|0BJV24C4SY]] satisfies $\\abs{\\bar z} = \\abs{z}$ and $\\abs{\\bar z - 1} = \\abs{z - 1}$, by the [[modulus of the conjugate|00KRCVE7CK]]. So $\\bar z$ lies on the circles centred $0$ and $1$ through $z$, and $\\bar z \\in \\mathcal{C}$.\n9. Let $z = x + iy \\in \\mathcal{C}$ with $x, y$ real. If $z$ is not real, then $x$ is the midpoint of $z$ and $\\bar z$, and $iy$ is the midpoint of $0$ and $z - \\bar z$. So $x, iy \\in \\mathcal{C}$. If $z$ is real, then $x = z$ and $iy = 0$.\n10. Let $y \\neq 0$ be real. If $y \\in \\mathcal{C}$, the circle centred $0$ through $y$ meets the imaginary axis at $iy$. If $iy \\in \\mathcal{C}$, the circle centred $0$ through $iy$ meets the real axis at $y$. So $y \\in \\mathcal{C}$ if and only if $iy \\in \\mathcal{C}$.\n11. By steps 9 and 10 and the sum $x + iy$ of step 5: $z \\in \\mathcal{C}$ if and only if $\\operatorname{Re} z \\in \\mathcal{C}$ and $\\operatorname{Im} z \\in \\mathcal{C}$.\n12. Let $x, y$ be nonzero reals in $\\mathcal{C}$. The point $x + iy - 1$ lies in $\\mathcal{C}$, so we can draw the line $\\set{x + t(iy - 1) : t \\in \\R}$ through $x$ parallel to the line through $1$ and $iy$.\n13. This line meets the imaginary axis where $x - t = 0$, at the point $ixy$. So $ixy \\in \\mathcal{C}$, and $xy \\in \\mathcal{C}$ by step 10. If $x = 0$ or $y = 0$, then $xy = 0 \\in \\mathcal{C}$.\n14. The line $\\set{i + t(1 - iy) : t \\in \\R}$ through $i$ and $i + 1 - iy$ meets the real axis where $1 - ty = 0$, at $1/y$. So $1/y \\in \\mathcal{C}$.\n15. Let $z = x + iy$ and $w = u + iv$ be in $\\mathcal{C}$, with $x, y, u, v$ real. Then $zw = (xu - yv) + i(xv + yu)$, so $zw \\in \\mathcal{C}$ by steps 6, 11 and 13.\n16. If $z \\neq 0$, then $r = x^2 + y^2 \u003e 0$ lies in $\\mathcal{C}$, and $1/z = x/r - i y/r$. So $1/z \\in \\mathcal{C}$ by steps 11, 13 and 14.\n17. By steps 6, 15 and 16, $\\mathcal{C}$ is a [[subfield|07QY5RPFBR]] of the [[field $\\C$|0FRSR10DWF]].\n18. Let $r \u003e 0$ be real in $\\mathcal{C}$. The midpoint $c = (r - 1)/2$ of $-1$ and $r$ is in $\\mathcal{C}$. The circle centred $c$ through $r$ has radius $(r+1)/2$.\n19. A point $iy$ with $y$ real lies on this circle if and only if $c^2 + y^2 = (r+1)^2/4$, that is, $y^2 = r$. With $y = \\sqrt r$ from the [[real square root|068BZBXE0C]], $i \\sqrt r \\in \\mathcal{C}$, so $\\sqrt r \\in \\mathcal{C}$.\n20. Let $z \\in \\mathcal{C}$ with $z \\neq 0$. By steps 8 and 17, [[$z \\bar z = \\abs{z}^2$|0ZB7J071WQ]] lies in $\\mathcal{C}$, so the [[modulus|0KW2X0TFBX]] $\\rho = \\abs{z}$ lies in $\\mathcal{C}$ by step 19. Let $w = z / \\rho$, so $\\abs{w} = 1$ and $\\bar w = 1/w$.\n21. If $w = -1$, then $(i \\sqrt \\rho)^2 = -\\rho = z$, and $i \\sqrt \\rho \\in \\mathcal{C}$.\n22. If $w \\neq -1$, let $s = \\sqrt \\rho \\, (w + 1)/\\abs{w + 1}$, which lies in $\\mathcal{C}$ by steps 17, 19 and 20. Then $s^2 = \\rho (w+1)^2 / ((w+1)(\\bar w + 1)) = \\rho (w + 1)/(\\bar w + 1)$.\n23. Since $\\bar w + 1 = (1 + w)/w$, this gives $s^2 = \\rho w = z$. So $z$ has a square root in $\\mathcal{C}$. Its negative, the other square root, is in $\\mathcal{C}$ too. For $z = 0$ the only square root is $0$.\n24. So $\\mathcal{C}$ is a subfield of $\\C$ that contains every square root of each of its elements.\n\nPart 2: minimality.\n\n25. Let $M$ be the intersection of all subfields of $\\C$ that contain every square root of each of their elements. Then $M$ is a subfield of $\\C$ with the same property.\n26. Conjugation preserves sums and products, and it is its own inverse. So $\\bar M = \\set{\\bar z : z \\in M}$ is a subfield of $\\C$. If $s^2 = \\bar z$ with $z \\in M$, then $\\bar s^2 = z$, so $\\bar s \\in M$ and $s \\in \\bar M$.\n27. So $\\bar M$ has the same property, and $M \\subseteq \\bar M$. Conjugating gives $\\bar M \\subseteq M$. So $M$ is closed under conjugation.\n28. The number $i$ is a square root of $-1 \\in M$, so $i \\in M$. For $z \\in M$, then $\\operatorname{Re} z = (z + \\bar z)/2$ and $\\operatorname{Im} z = (z - \\bar z)/(2i)$ lie in $M$. Let $M_\\R = M \\cap \\R$.\n29. Every real $r \\ge 0$ in $M_\\R$ has $\\sqrt r \\in M_\\R$. A line through two points of $M$ has an equation $\\alpha x + \\beta y = \\gamma$ with $\\alpha, \\beta, \\gamma \\in M_\\R$, not both $\\alpha, \\beta$ zero. Here $z = x + iy$. For the points $a_1 + i a_2$ and $b_1 + i b_2$, take $\\alpha = b_2 - a_2$, $\\beta = a_1 - b_1$ and $\\gamma = \\alpha a_1 + \\beta a_2$.\n30. A circle centred at $c_1 + i c_2 \\in M$ through a point of $M$ has an equation $x^2 + y^2 + \\delta x + \\epsilon y + \\eta = 0$ with $\\delta = -2c_1$, $\\epsilon = -2c_2$ and $\\eta \\in M_\\R$.\n31. Two distinct lines that meet have a nonzero determinant $\\alpha \\beta' - \\alpha' \\beta$. By Cramer's rule, the common point has coordinates in $M_\\R$.\n32. For a line and a circle, say $\\beta \\neq 0$; otherwise swap the roles of $x$ and $y$. Substituting $y = (\\gamma - \\alpha x)/\\beta$ gives $A x^2 + B x + C = 0$ with $A = 1 + \\alpha^2/\\beta^2 \\neq 0$ and $B, C \\in M_\\R$.\n33. A real solution $x$ needs $D = B^2 - 4AC \\ge 0$, and then [[$x = (-B \\pm \\sqrt D)/(2A)$|0KYZXYDGQZ]]. So $x \\in M_\\R$, and $y \\in M_\\R$.\n34. For two distinct circles, subtracting their equations gives $(\\delta - \\delta')x + (\\epsilon - \\epsilon')y + (\\eta - \\eta') = 0$. If $\\delta = \\delta'$ and $\\epsilon = \\epsilon'$, the circles are concentric and distinct, so they have no common point.\n35. Otherwise, the common points are the common points of the first circle with this line. They lie in $M$ by step 33.\n36. So each step of a construction adds a point $x + iy$ with $x, y \\in M_\\R$, which lies in $M$ since $i \\in M$. As $0, 1 \\in M$, induction on the number of steps gives $\\mathcal{C} \\subseteq M$.\n37. Every subfield $K$ of $\\C$ closed under square roots contains $M$, so $\\mathcal{C} \\subseteq K$. With step 24, $\\mathcal{C}$ is the smallest subfield of $\\C$ closed under square roots.\n\n## Idea\n\nMidpoints, parallels and similar triangles give sums, products and quotients, and a circle on a diameter of length $1 + r$ gives $\\sqrt r$. A complex square root needs only a real square root and the bisector direction $w + 1$. Conversely, the common points of lines and circles solve linear or quadratic equations, so each step needs only field operations and one square root.\n",
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