Doubling the cube

The Egyptians, Indians, and particularly the Greeks^[2] were aware of the problem and made many futile attempts at solving what they saw as an obstinate but soluble problem.^[3][b] However, the nonexistence of a compass-and-straightedge solution was finally proven by Pierre Wantzel in 1837.


In algebraic terms, doubling a unit cube requires the construction of a line segment of length x, where x³ = 2; in other words, x = ${\displaystyle {\sqrt[{3}]{2}}}$, the cube root of two. This is because a cube of side length 1 has a volume of 1³ = 1, and a cube of twice that volume (a volume of 2) has a side length of the cube root of 2. The impossibility of doubling the cube is therefore equivalent to the statement that ${\displaystyle {\sqrt[{3}]{2}}}$ is not a constructible number. This is a consequence of the fact that the coordinates of a new point constructed by a compass and straightedge are roots of polynomials over the field generated by the coordinates of previous points, of no greater degree than a quadratic. This implies that the degree of the field extension generated by a constructible point must be a power of 2. The field extension generated by ${\displaystyle {\sqrt[{3}]{2}}}$, however, is of degree 3.

Proof of impossibility[edit]

We begin with the unit line segment defined by points (0,0) and (1,0) in the plane. We are required to construct a line segment defined by two points separated by a distance of ${\displaystyle {\sqrt[{3}]{2}}}$. It is easily shown that compass and straightedge constructions would allow such a line segment to be freely moved to touch the origin, parallel with the unit line segment - so equivalently we may consider the task of constructing a line segment from (0,0) to (${\displaystyle {\sqrt[{3}]{2}}}$, 0), which entails constructing the point (${\displaystyle {\sqrt[{3}]{2}}}$, 0).


Respectively, the tools of a compass and straightedge allow us to create circles centred on one previously defined point and passing through another, and to create lines passing through two previously defined points. Any newly defined point either arises as the result of the intersection of two such circles, as the intersection of a circle and a line, or as the intersection of two lines. An exercise of elementary analytic geometry shows that in all three cases, both the x- and y-coordinates of the newly defined point satisfy a polynomial of degree no higher than a quadratic, with coefficients that are additions, subtractions, multiplications, and divisions involving the coordinates of the previously defined points (and rational numbers). Restated in more abstract terminology, the new x- and y-coordinates have minimal polynomials of degree at most 2 over the subfield of ${\displaystyle \mathbb {R} }$ generated by the previous coordinates. Therefore, the degree of the field extension corresponding to each new coordinate is 2 or 1.


So, given a coordinate of any constructed point, we may proceed inductively backwards through the x- and y-coordinates of the points in the order that they were defined until we reach the original pair of points (0,0) and (1,0). As every field extension has degree 2 or 1, and as the field extension over ${\displaystyle \mathbb {Q} }$ of the coordinates of the original pair of points is clearly of degree 1, it follows from the tower rule that the degree of the field extension over ${\displaystyle \mathbb {Q} }$ of any coordinate of a constructed point is a power of 2.


Now, p(x) = x³ − 2 = 0 is easily seen to be irreducible over ${\displaystyle \mathbb {Z} }$ – any factorisation would involve a linear factor (x − k) for some k ∈ ${\displaystyle \mathbb {Z} }$, and so k must be a root of p(x); but also k must divide 2 (by the rational root theorem); that is, k = 1, 2, −1 or −2, and none of these are roots of p(x). By Gauss's Lemma, p(x) is also irreducible over ${\displaystyle \mathbb {Q} }$, and is thus a minimal polynomial over ${\displaystyle \mathbb {Q} }$ for ${\displaystyle {\sqrt[{3}]{2}}}$. The field extension ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}):\mathbb {Q} }$ is therefore of degree 3. But this is not a power of 2, so by the above, ${\displaystyle {\sqrt[{3}]{2}}}$ is not the coordinate of a constructible point, and thus a line segment of ${\displaystyle {\sqrt[{3}]{2}}}$ cannot be constructed, and the cube cannot be doubled.