I gotta see that!

It doesn't easily follow. How did you do it?

​@@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

​@@vekyllBoth could be negative so (i) doesn't prove that at least one of them is positive.

I hope you found it as interesting as I did!

Very much so! I love KZfaq channels that take viewer requests like that!

@@jackhimes4400 Well, when I know how to do it 😅

Because the previous line says that -i>0. You only flip the inequality if you multiply by a negative number.

@@snellbrosmath oops gotcha that makes sense.

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.

I think total order takes on additional meaning in the context of rings and field, but this might be a linear order actually

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) 👍

@@swenji9113 is it? Well, it's been a while since I've messed with that part of set theory.

@@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 :)

The problem comes from point iii. RxR can be totally ordered lexicographically. It can even be well ordered, using the axiom of choice…

But it is not possible to define a total order on any field. Actually, R is the only totally ordered uncountable field.

@@canaDavid1 Sure, that's why it would be better to have a clearer title

Sorry :)

​@@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...

Can you elaborate?

Could you elaborate?

> 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.

What definition? i=sqrt{-1}?

Thank you!

No, why do you ask?

Glad it helped you!!!

The author of this book decided to give this definition for you to show how complex numbers "compare" with each other.

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.