# Ordered Fields

Michael Redman

Updated 2019 May 19

# Definition of Ordered Field

A field F is an ordered field if it contains a subset P such that:

Closure under Addition.If x and y are in P, then x+y is in P.

Closure under Multiplication.If x and y are in P, then xy is in P.

Total Ordering. for any x in F, exactly one of these is true: x is in P, -x is in P, or x=0.

The subset P is called the positive set of F

# Definition of < and >

If x and y are in an ordered field with a positive set P, then x<y means y-x is in P.

# 1 $$\in$$ P, which by definition of > also means 1>0

By trichotomy axiom, exactly one of 1 or -1 is in P. However by field properties (see page on fields), $$(-1)^2$$=1, so by clousre of P under multiplication, if (-1) in P then 1 also in P, which is a contradiction.

This also explains why the 2-element field consisting of 0 and 1, cannot be an ordered field.

# If x<0 and y>0 then xy<0

x<0 means -x in P. Then by closure under multiplication -xy in P, and by total ordering property -(-xy)=xy cannot also be in P. Since neither x nor y is 0 xy must be less than 0.

(This uses the properties that in a group -(-x)=x and in a field xy=0 only if at least one of x or y is 0. See the "groups" and "fields" docs for proofs of those propositions.)

# x not 0 implies $$x^2$$ > 0

If x>0 then true by closure under multiplication of the positive set.

If x<0 then by the total ordering property -x in P, and then by closure of the positive set so is $$(-x)^2$$. Since in any field -x = (-1)*x and $$(-1)^2$$=1 (see "fields" doc), $$x^2=(-x)^2$$ so $$x^2$$ in P.

# 0$$\leq$$x<y and 0$$\leq$$u<v implies xu<yv

yv is always positive by closure of the positive set. If either x or u is 0, then xu is 0 and yv is greater.

If neither x nor u is zero, then by closure of the positive set under multiplication, (y-x)(v-u)=yv-uy-xv+ux>0, and u(y-x)=uy-ux>0, and x(v-u)=vx-ux>0. By closure of the positive set under addition the sum of the left side quantities, all positive, is also positive; this is yv-ux.