Your first sentence and what comes after "In other words" are very different. "A total order" and "A total order compatible with the ring structure" are two very different things... By the well-ordering "axiom", C is even well-orderable, which is a lot stronger statement than just totally orderable
A much easier proof is possible: first show that for any nonzero x, exactly one of x and -x is positive (easily follows from ii), and in both cases x² is positive (easily follows from iii and i). Then both 1=(-1)² and -1=i² must be positive, a contradiction.
@@irhzuf in (ii), set (z1, z2, z3) = (x, 0, -x). That is, suppose x > 0 so that x + (-x) > 0 + (-x), implying that 0 > -x or -x is negative. Alternatively, set (z1, z2, z3) = (0, x, -x) so that -x is positive and x is negative. By (i), one of these cases must be hold if x ≠ 0.
@@DrakePitts maybe easier: "at least one" (of x and -x is positive) follows from (i). Now if both x and -x were positive, x+(-x) would be greater than x+0, that is, 0>x, a contradiction.
@@DrakePittsThanks! And I guess that in (iii) if x>0 then you pick (x,0,x) so x^2>0 from (i) it can't be different. (a similar thing can be done if -x>0) Then if we assume that 1>0 then -1
How widespread is this definition of total order? I thought the standard meaning of total order is just about comparisons, without arithmetic requirements, in the same way that partial orders are
I think he meant to say that C cannot be an ordered field. Any _set_ can be totally ordered (well, if you accept accept the well ordering axiom), but if you are ordering an algebraic structure of some kind you want the ordering to be preserved under the operation(s). You can also look at finite fields and see why they cannot be ordered, even though it is trivial to order their elements (independent of the operations).
Actually, the fact that every set has a linear order is independent from ZF but it's stricly weaker that the well-ordering axiom (equivalently the axiom of choice) 👍
@@Phylaetra Yes, I remember proving that any set can be linearly ordered from the completeness theorem, which is equivalent to the ultrafilter lemma. I know nothing about the proof but I know the ultrafilter lemma is strictly weaker than AC :)
@@canaDavid1 Any subfield of R which is uncountable is also totally orderable as a field... Any field where -1 is not a sum of squares is orderable as a field...
> is just a relation, not necessarily the "greater than" relation that we are accustomed to. That's why we need to get a contradiction in the way we did.
Another note, transitive, requires that elements have a relationship (are comparable) in order to show transitivity (a>b and b>c => a>c). If the elements aren't comparable, then we can not fulfill the assumption for transitive.
TLDR 0=i leads to every complex number having the same size I'm unfamiliar with orderings, so I may have a stupid question: why can we say 0!=i? Of course we know this from our usual use of the equals sign, but aren't we constructing a new relation to order C that may be very different from the usual "="? Do we technically need to show it out, like maybe the extra step of squaring each side of 0=i (maybe repeatedly to get 0=-1 and 0=1) and observing that our additive and multiplicative identities being the same perhaps violates having axioms 2 and 3 simultaneously true?
I'm thinking aloud here. If 0=i we also have 0=i=1=-1. This immediately makes our concept of bigger and smaller complex numbers different from that of the real numbers which is unfortunate but doesn't mean an ordering isn't possible for C. Let's play with axiom 2. Say we have two complex numbers z and w and z>w. By axiom 2, z+0>w+0. Our 0=i relation will give us a few extra inequalities, eg z-1>w+1. We can again add zero to both sides of this inequality and inductively find an infinite set of inequalities, z+n+mi>w+k+qi, where n,m,k, and q are integers. We have shown that any complex number y=w+k+qi (with k and q integers) satisfies the same inequalities as w. We certainly couldn't have yw then because it would produce contradictions, so y=w if 0=i
One minute lol, I should have done this first: let z be a complex number. In the ordinary sense, z = (-1)×(-1)×z. In our ordering sense we know -1=0, so we get z= 0×(-1)×z=0. Therefore every complex number has the same size as 0. So letting 0=i gave us a trivial ordering. Wouldn't that make 0=i valid in some sense?
In general, we say that 0 cannot equal 1 in a ring. Letting 0=i implies that 0=1, and therefore invalidated 0 from being the multiplicative identity, and Vice versa.
