Duplication of the cube-one of the fundamental problems of Greek geometry, together with squaring the circle and trisecting the angle. It asks whether, given a cube of a certain size, it is possible to construct a cube of double the original size, using only a compass and a straightedge. Although the Greeks came up with several ingenious methods for doubling the cube, they were never able to do so without resorting to additional instruments. It is, in fact, impossible to double the cube with only a compass and a straightedge, but a proof of this fact did not come until more than two thousand years after the problem was first posed.
There are several different stories about the origin of the cube-doubling problem. One story, mythological in origin, says that King Minos of Crete ordered a tomb built for his son Glaucus, who died by drowning in a jar of honey. But when the tomb was finished, Minos said, "Too small is the tomb you have marked out as the royal resting place. Let it be twice as large. Without spoiling the form, quickly double each side of the tomb." The poet who related this story may have been a fine wordsmith, but he was far from an able mathematician, since doubling each dimension of the tomb would produce a tomb that is 8 times as large as the original, not 2 times.
According to another story, when the Delians were suffering from a terrible plague, the oracle told them that they would be spared if they constructed an altar that was double the size of the existing one. Eratosthenes, as quoted by Theon of Smyrna, wrote that "Their craftsmen fell into a great perplexity in their efforts to discover how a solid could be made the double of a similar solid; they therefore went to ask Plato about it, and he replied that the oracle meant, not that the god wanted an altar of double the size, but that he wished, in setting them the task, to shame the Greeks for their neglect of mathematics and their contempt of geometry."
The first genuine progress in solving the problem was made by Hippocrates, who realized, using purely geometric considerations, that the cube-doubling problem could be solved if it were possible to find two mean proportionals between 1 and 2. Following Hippocrates's insight, many of the best Greek mathematicians came up with clever mechanical solutions to the problem. Eratosthenes was one of those who devised a mechanical method, and in fact he built a column at Alexandria for King Ptolemy, with his cube-doubling method inscribed on it. He suggested that Ptolemy always adopt his method, and warned him against his rivals' methods: "Do not thou seek to do the difficult business of Archytas's cylinders, or to cut the cone in the triads of Menaechmus, or to compass such a curved form of lines as is described by the god-fearing Eudoxus.... Thus may it be, and let any one who sees this offering say 'This is the gift of Eratosthenes of Cyrene.'"
The Greeks probably suspected that the problem of doubling the cube using compass and straightedge was impossible. But this was not proved until 1837, when Pierre Wantzel used algebraic techniques to prove the impossibility not only of doubling the cube, but also of squaring the circle and trisecting the angle.
The key idea in this algebraic method was to turn the problem into a question about numbers, by asking what kind of lengths can be constructed using straightedge and compass, starting with a unit length. For example, 2 is a constructible number, since a line segment of length 2 can be constructed simply by building two unit-segments that abut each other. Once mathematicians had made the definition of constructible numbers, they could ask what kind of equations a constructible number could satisfy. They discovered a remarkable fact: no constructible number could satisfy a cubic polynomial equation, like x[sup3 ]-5=0 (unless the polynomial was of a very simple type called reducible).
If you start with a cube whose dimensions each measure one unit in length, then doubling the cube amounts to building a cube whose dimensions have length equal to the cube root of 2 (so that their product, the volume of the cube, is equal to 2). So the question of doubling the cube reduces to the question, is the cube root of 2 a constructible length? The cube root of 2 is a solution to the irreducible cubic equation x[sup3 ]-2=0, so by the algebraic theorem, it cannot be a constructible number! Hence it is impossible to double the cube with only a compass and straightedge.
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