A field is "ordered" if its nonzero elements can be split into two sets, P and N, such that P contains x iff N contains -x, and P is closed under addition and multiplication.
Use these sets to define a < relation. Assert x < y iff y-x is in P. Show that this relation is irreflexive, antisymmetric, and transitive. That is, x-x is 0 (not in P), x-y has the opposite sign of y-x, and if y-x and z-y are in P then so is z-x, P being closed under addition, so x < z as expected. The elements of the field are linearly ordered.
Show that a < b & c < d implies a+c < b+d, and a < b & 0 < c implies ac < bc. These follow from the fact that P is closed under addition and multiplication.
Since -1 and -1×-1 cannot both be positive, -1 is negative, and 1 is positive. Successive integers are all positive. If the field has nonzero characteristic, successive integers eventually reach -1, whence -1 is positive, which is impossible. Every ordered field has characteristic 0. The integers go on forever. Remember that a finite cycle, x 2x 3x … kx = 0, means k is the same as 0. Thus an ordered field cannot have a finite additive cycle.
The rationals and the reals are ordered, but complex numbers cannot be, since neither i nor -i can be positive, their squares being -1.
There is no nontrivial automorphism of the rationals, since 0 and 1 map to themselves, and that determines everything else. We will use the properties of an ordered field to show the reals have no nontrivial automorphisms either. Squares map to squares under an automorphism, and in the case of the reals, a number is a square iff it is positive. Thus an automorphism maps P onto P and N onto N. We already know it preserves the rationals. Let an automorphism move x, and consider two rationals on either side of x. The rationals remain fixed, and x stays between these rationals. This holds for arbitrarily small intervals, so x maps to x.