A number a is constructible
if we can reach it by a finite number of geometric constructions.
If two numbers a and b are constructible then a+b, a/b, and ab are also
constructible. This implies that the set, C, of all
constructible numbers is a field.
Further we have already shown that the field of rational
numbers, Q, is a subset of the constructible numbers.
We can imagine building C up by starting with the rational
numbers and then adjoining each new
element that we find one by one, defining intermediate fields, Fn,
along the way (by "new" I mean an element that we did not previously have
in our field.) e.g.
Q=F1 < F2 < ... < Fn < Fn+1< CWe must always construct points using a straight-edge and compass, so each intersection of lines and/or circles is a potentially new constructible point. Since the algebraic form of a constructed line is: ax + by + c = 0where a,b,and c are all previously constructed numbers, and the algebraic form of a constructed circle is: x2+y2+ax+by+c = 0where a,b, and c are all previously constructed numbers, the solution of any intersection point will always be either an element of the current field, Fn, or it will be of the form: x =Which implies that x is then an elements of Fn( 
) where a=B2-4AC, an element of Fn. This implies that the form of the
intermediate fields, Fn, can always be written as a series of
single element adjoins. i.e.,
Q=F1 < F2 < ... < Fn < Ccan now be written as Q=F < F(a1) < F(a1,a2) < ... < F(a1,...,an) < C.In each case the adjoined element was either in the previous field, or is new implying that the degree of of each succesive field over its predecessor is either 1 or 2. [F(a1,...,an):F(a1,...,an-1)] = 1or 2,therefore [F(a1,...,an):Q] = 2m. |
