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From my textbook,

The definition of the order relation denoted by > in the real number system is based upon the existence of a subset $P$ (the positive reals) having the following properties.

i) For any number $a\ne0$, either $a$ or $-a$ but not both belong to $P$

ii)If $a$ and $b$ belong to $P$, so does $a+b$

iii)If $a$ and $b$ belong to $P$, so does $ab$

When such a set $P$ exists we write $a>b$ if and only if $a-b$ belongs to $P$.

Prove that the complex number system does not possess a nonempty subset $P$ having properties i), ii) and iii).

The answer: Assume $i$ is in $P$ $\implies i^2=-1\in P\implies(-1)i\implies-i \in P$, which violates i). And similarly for $-i$

My question: This proves that $i$ and $-i$ cannot be in $P$, but I have trouble showing that a general complex number $a+bi$ cannot be in $P$

matt
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    Why do you need to do that? You only need to show that such a $P$ doesn't exist, and it seems you've done that. – Brian Tung Aug 19 '21 at 19:58
  • Can't I have a $P$ with $a+bi$, which satisfies all three properties? – matt Aug 19 '21 at 20:00
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    Note that property (i) doesn't merely need to be satisfied for a given number; it has to be satisfied for all numbers $\not= 0$: Either the number has to be in $P$, or its additive inverse is, but not both. You've shown that there is a number (namely, $i$) for which $P$ does not pass condition (i). – Brian Tung Aug 19 '21 at 20:03
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    Otherwise, the positive real integers would qualify: No negative real integer is present, and they are closed over addition and multiplication. But clearly they don't suffice to represent sign in the complex field! – Brian Tung Aug 19 '21 at 20:06
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    @matt This set $P$ has to have all three properties, including the first property which should work for every complex number, including $i$. The proof shows that, if the third property also holds, then the first property does not, since the first property does not hold for every complex number. It's possible to define certain sets where the first property only holds for certain complex numbers (e.g. if $P$ literally is the positive reals, then ii and iii both hold, and i holds only for real numbers in $\Bbb{C}$), but this doesn't properly satisfy i. – Theo Bendit Aug 19 '21 at 20:07
  • Thank you. The properties need to be satisfied for all numbers. That makes so much more sense. – matt Aug 19 '21 at 20:11

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