To see this, assume that a field F contains only constructible numbers, and imagine a Cartesian plane where all our constructions take place. To sum up, let there be a constructible number a . This simply means that the number was determined through a sequence of geometric constructions with a ruler and a compass. This state of affairs lasts until we attempt to determine points of intersection with a circle. The fields, if necessary, extend until the number a crops up in the course of the construction.
Every constructible number is algebraic .
Continuing this process we shall eventually come up with a polynomial equation with rational coefficients satisfied by a . This proves a is algebraic. Since the set of all algebraic numbers is countable , we get, as an immediate consequence of this result, that the set of all constructible numbers is also countable. Not all algebraic numbers are constructible. As we shall see later on, this gives a negative solution to the problem of Doubling the Cube, one of the three famous problems of antiquity.