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Saff and Snider in their book define order axioms as below:

  1. If $\alpha\neq 0$ then either $\alpha>0$ or $\alpha<0$.
  2. If $\alpha>0$ and $\beta>0$ then $\alpha+\beta>0$.
  3. If $\alpha>0$ and $\beta>0$ then $\alpha\beta>0$.

I have to prove the impossibility of ordering in $\mathbb{C}$. So suppose $i>0$. Then using (3), $$i^2>0\Rightarrow -1>0\Rightarrow (-1)(-1)>0\Rightarrow 1>0.$$ Now I want to show that $0>1$ as well to get a contradiction. How do I proceed?

matrixx
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  • Apply (2) and get the desired contradiction. –  Apr 13 '17 at 03:47
  • Actually, given your axioms, it doesn't. But your axioms fail to prove that any exists that is greater than 0. Most texts give the axis that if a <b then a+x < b+x for all x. I assumed you had that as an axiom. With your axioms, it is possible that i < 0, 1 <0, and -1 <0. Although it is impossible that i >0 or that -1>0 (else -1.-1=1>0 and 1+-1=0>0). But there's nothing wrong with i, 1,and -1 all less then zero. – fleablood Apr 14 '17 at 07:39

1 Answers1

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If $i>0$ then $i^2=-1>0$.

Then $-1*-1 =1 >0$

Then $-1+1=0 >0$

A contradiction.

If $i <0$ then $0- i >0$ (def. $a>b\iff a-b >0$).

Then $-i*-i =i^2= -1>0$ and $-1*-1=1 >0$ and $-1+1=0>0$. A contradiction.

===

First prove 1) if $x>0$ then $-x <0$ and vise versa.

Then prove 2) $x^2 > 0$ for all $x \ne 0$.

Then conclude $i^2=-1 >0$ and $1^2=1>0$ which contridicts 1).

Proof of 1) if $0 > -x \iff 0-(-x)=x >0$

Proof of 2) if $x > 0$ then $x^2= x*x>x*0=0$

If $x <0$ then $-x >0$ so $(-x)(-x)>0$ and $(-x)(-x)=x^2$.

(Of course, if you are really picky, you have to also prove $(-a)(-b)=ab $.)

fleablood
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  • In proof of (1) how does $x+(-x)>0+(-x)$ follow from $x>0$? – matrixx Apr 14 '17 at 07:10
  • Actually, given your axioms, it doesn't. But your axioms fail to prove that any exists that is greater than 0. Most texts give the axis that if a <b then a+x < b+x for all x. I assumed you had that as an axiom. With your axioms, it is possible that i < 0, 1 <0, and -1 <0. Although it is impossible that i >0 or that -1>0 (else -1.-1=1>0 and 1+-1=0>0). But there's nothing wrong with i, 1,and -1 all less then zero. – fleablood Apr 14 '17 at 07:39
  • yes my confusion came from this second axiom. – matrixx Apr 14 '17 at 07:46
  • I have never seen the axioms presented this way. I think they could work but we need... more. There's no way to compare if a >0 and b>0 then how do a and b compare? Nor how can you determine that anything is ever positive. The more I think of it these axioms are just not right. – fleablood Apr 14 '17 at 07:55
  • It does say on a separate paragraph that $\alpha>\beta$ if and only if $\alpha-\beta>0$. Does that help? – matrixx Apr 14 '17 at 08:04
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    That helps immensely – fleablood Apr 14 '17 at 08:06
  • So a > b => a-b > 0 => a-b+x-x > 0 => a+x > b+x for all x. And a+x > b+x for any x => a+x-b-x >0 => a-b>0 => a>b. – fleablood Apr 14 '17 at 08:17
  • So $\alpha>\beta\Leftrightarrow \alpha-\beta>0$ should be on the axioms list. – matrixx Apr 14 '17 at 08:23
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    Yes, it should. Or it should be a later definition. $x >0$ is an abstract concept; but $x > y $ is defined as $x-y > 0$. – fleablood Apr 14 '17 at 08:26
  • I think it's not an axiom, it's a definition of what $\alpha>\beta$ (as opposed to $\alpha>0$) signifies. – ancient mathematician Apr 14 '17 at 08:26
  • Ancientmathematician, exactly. – fleablood Apr 14 '17 at 08:27
  • understood, thanks :) – matrixx Apr 14 '17 at 08:30
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    So there is an abstract concept that every non zero number has some order compared to 0 so that either x>0 or x <0. Then there are axioms given. Then a definition for order between nonzero numbers is given. Then a bunch of propositions can be proven via axioms. – fleablood Apr 14 '17 at 08:31