Re: 1+i > i

From: William Elliot (marsh_at_privacy.net)
Date: 08/15/04 

Date: Sun, 15 Aug 2004 03:10:44 -0700

From: Stephen J. Herschkorn <herschko@rutcor.rutgers.edu>
Newsgroups: sci.math
Subject: Re: 1+i > i

>
>In an ordered field, a > b implies a + c > b + c; b positive
>implies a + b > a, and a - b > 0 implies a > b.
>Problem is, the field of complex numbers is not an ordered field,
>though its subfield of reals is.

Indeed, there are no total orders for the ring of complex numbers.
However there is a unique partial order for C extending the order of R.
The axioms for an ordered rings are (R,<=,+,*), is (R,+,*) is a ring,
(R,<=) is an (partially) ordered set, <= is translation invariant, ie
        a <= b ==> a + c <= a + c
the ordered group axiom, and finally the ordered ring axiom
        0 <= a,b ==> 0 <= ab
or equivalently
        a <= b, 0 <= c ==> ac <= bc

Now for an (partially) ordered field, are other axioms needed, like
        0 < a ==> 0 < 1/a
and/or
        1 <= a ==> 1/a <= 1 ?

Can fields be partially ordered? Yes for example, the unique partial
order for the ring C extending the order of R, suffices for the field C
including the ordered field axioms above. (See my first post in this
thread for details of the unique order.)

If 1 <= a, then if a = 1, 1/a = 1 <= 1 and if 1 < a, then
        0 < a - 1; 0 < 1/(a - 1); 0 <= (a - 1)/(a - 1) = 1
        0 <= 1 < a; 0 < a; 0 < 1/a; 1/a <= a/a = 1
Thus
        1 <= a ==> 1/a <= 1
need not be an axiom.

If the field is totally ordered, then if 0 < a and not 0 < 1/a
comes contradiction
        1/a <= 0; 0 <= -1/a; 0 <= -a/a = -1; 0 <= -1
        1 <= 0 <= (-1)^2 = 1; 0 = 1
Thus for totally ordered fields
        0 < a ==> 0 < 1/a
is not needed as an axiom.

Anyway, the axiom 0 < a ==> 0 < 1/a shows 0 < 1 for any non-trivially
ordered field. As the order is non-trivial, some a,b with
        a < b; 0 < b - a = c
        0 < 1/c; 0 <= c/c = 1
and since 0 /= 1 by definition of field, 0 < 1
        Hence it's observed a non-trivially ordered field has
characteristic 0 and that the prime subfield generate by 1, is Q with the
usual total order. 

----

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