2

Say we are given this:

Impossibility of ordering the complex numbers. As yet we have not defined a relation of the form $x < y$ if $x$ and $y$ are arbitrary complex numbers, for the reason that it is impossible to give a definition of $<$ for complex numbers which will have all the properties in Axioms 6 through 8.

To illustrate, suppose we were able to define an order relation $<$ satisfying Axioms 6, 7, and 8. Then, since $i \neq 0$, we must have either $i > 0$ or $i < 0$, by Axiom 6. Let us assume $i > 0$. Then, taking $x = y = i$ in Axiom 8, we get $i^2 > 0$, or $—1 > 0$. Adding 1 to both sides (Axiom 7), we get $0 > 1$. On the other hand, applying Axiom 8 to $—1 > 0$ yields $1 > 0$. Thus we have both $0 > 1$ and $1 > 0$, which, by Axiom 6, is impossible. Hence the assumption $i > 0$ leads us to a contradiction.

So we are given this "passage" and the question is

By reading the passage: Suppose that $<$ is a relation on $\mathbb{C}$ that satisfies Axioms 6, 7, and 8. and show that the assumption $i <0$ leads to a contradiction.

Axiom 6 - Exactly one of the relations $x = y$, $x<y$ and $x>y$ holds. Note $x>y$ means the same thing as $y<x$

Axiom 7 - If $x<y$, then for every $z$ we have $x+z <y+z$.

Axiom 8 - If $x>0$ and $y>0$, then $xy>0$.

So this is what I have but I am really confused from that passage. Didn't we already complete the proof because we ended up with a contradiction? Help on this one please.

kahen
  • 15,760
Carlo Sanchez
  • 83
  • How do you get $i^2\gt0$ from Axiom 8? – Gerry Myerson Mar 08 '13 at 00:43
  • @Gerry: Axiom $8$ plus the assumption that $i>0$. – Brian M. Scott Mar 08 '13 at 00:44
  • Axiom 8 doesn't mention multiplication. – Peter Shor Mar 08 '13 at 00:44
  • Yes, multiplication should be also covered by the axioms.. – Berci Mar 08 '13 at 00:45
  • Carlo, you have only half the proof there: the passage shows that the assumption that $i>0$ leads to a contradiction. Now you’re to show that the assumption that $i<0$ also leads to a contradiction. Since $i\ne 0$, that will show that no order on $\Bbb C$ can satisfy these axioms. – Brian M. Scott Mar 08 '13 at 00:45
  • 1
    You don't have a proof. Since the axioms don't mention multiplication, you can't conclude that $i>0$ implies $i^2 > 0$. – Peter Shor Mar 08 '13 at 00:46
  • 1
    @Peter: Axiom $8$ is misstated: it’s supposed to be that $x>0$ and $y>0$ implies that $xy>0$, as in this question. I expect that Carlo just typed the wrong axiom with the right number. – Brian M. Scott Mar 08 '13 at 00:47
  • What are the other axioms? Isn't it only a mistype? In the text it seems that the axiom for multiplication is used. – Berci Mar 08 '13 at 00:47
  • Have you quoted axiom 8 correctly? It doesn't seem to lead to the conclusions the existing text tries to draw from it. I suspect it was supposed to say something about the relation between ordering and multiplication, but as written here it just specifies that the ordering must be transitive. – hmakholm left over Monica Mar 08 '13 at 00:48
  • Carlo, there's a consensus building here that you've typoed Axiom 8. Please compare what you've written to what you meant to write, and edit accordingly. – Gerry Myerson Mar 08 '13 at 00:56
  • wow I am so sorry, I am loosing my mind I just edited it – Carlo Sanchez Mar 08 '13 at 01:00
  • A quick heads-up on mixing maths and text: I replaced the fragment $x=y,x<y,x>y$ with $x=y$, $x<y$ and $x>y$ because it had incorrect spacing. When writing LaTeX, punctuation goes outside $ .. $ but inside \[ .. \] (and friends). – kahen Mar 08 '13 at 01:17

1 Answers1

1

The passage proves only for the case when $i>0$ was considered, and the $i<0$ doesn't readily follow from this, though it is not hard neither:

By axiom 7., and if $i<0$, we have $0=i+(-i)<0+(-i)=-i$, that is, $-i>0$. But then the passage can be applied again, as $(-i)^2=i^2=-1$.

Berci
  • 90,745