Editing Doubling the cube (section)

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

Respectively, the tools of a compass and straightedge allow us to create [[circles]] [[centre (geometry)|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 {{mvar|x}}- and {{mvar|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 {{mvar|x}}- and {{mvar|y}}-coordinates have [[minimal polynomial (field theory)|minimal polynomial]]s of degree at most 2 over the [[Field (mathematics)|subfield]] of {{math|[[real number|<math>\mathbb{R}</math>]]}} generated by the previous coordinates. Therefore, the [[degree of a field extension|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 [[mathematical induction|inductively]] backwards through the {{mvar|x}}- and {{mvar|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 [[rational number|<math>\mathbb{Q}</math>]] of the coordinates of the original pair of points is clearly of degree 1, it follows from the [[Degree of a field extension#The multiplicitivity formula for degrees|tower rule]] that the degree of the field extension over <math>\mathbb{Q}</math> of any coordinate of a constructed point is a [[power of 2]].

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