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A straightedge and compass construction uses finitely many steps drawing lines through known points and circles with known centres and constructible radii, and adjoining their intersection points. The coordinates obtained by line–line intersections require only field operations; line–circle and circle–circle intersections require at most a quadratic extension after eliminating a linear equation. Thus all coordinates lie in a tower of quadratic extensions of the initial coordinate field. The degree of any finite intermediate coordinate field containing that initial field divides a power of two and is therefore $\boxed{2^r}$ for some $r\ge 0$.

For the source's notation this assumes $K\supseteq K_0$, as is customary when previously constructed points are retained. If only the coordinates of a final subset are called $K$, replace it by the compositum $K_{0}K$; otherwise the expression $[K:K_0]$ need not be defined. The power-of-two condition is necessary and is not being asserted as sufficient for an arbitrary algebraic number.

Starting from a unit cube, doubling its volume would construct a length $\sqrt[3]{2}$. Its polynomial $X^3-2$ is Eisenstein at two, so its degree over $\mathbb{Q}$ is three, which cannot divide a power of two. Thus doubling the cube is impossible by ruler and compass.

Set $c=\cos \theta$, $s=\sin \theta$. The real part of $(c+is{)}^5$ is $c^5-10c^3{s}^2+5c{s}^4$. Replacing $s^2$ by $1-c^2$ givesChoose the given angle to have cosine $5/6$. It is constructible, since its unit circle point has coordinates $(5/6,\sqrt{11}/6)$. If a fifth of that angle were constructible and had cosine $c$, the identity would implyThis polynomial is Eisenstein at five: all nonleading coefficients are divisible by five and the constant term is not divisible by 25, while 96 is not divisible by five. Hence $[\mathbb{Q}(c):\mathbb{Q}]=5$. But the initial angle field is $\mathbb{Q}(\sqrt{11})$, itself of degree two, so adjoining any finite collection of constructible coordinates to it would still give total degree over $\mathbb{Q}$ a power of two. It could not contain a degree-five element. Thus general angle quintisection is impossible with straightedge and compass.