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In an ordered field, must the multiplicative identity be positive? Or must it be defined as such?

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up vote 12 down vote accepted

Remember that for a total order,

  1. If $a \leq b$, then $a+c \leq b+c$.
  2. If $0 \leq a$ and $0 \leq b$, then $0 \leq ab$.

If $1 \leq 0$, then $1+(-1) \leq 0 + (-1)$ i.e. $0 \leq -1$. By ($2$), we need $0 \leq (-1)(-1) = 1$.

Hence, we get that $1 \leq 0 \leq 1$. For a non-trivial field, $0 \neq 1$. Hence, we get a contradiction that $$1 < 0 < 1.$$

Hence, $0 < 1$.

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Minor point - $0 = 1$ is excluded by definition of field. – Maciej Piechotka Jun 3 '12 at 5:19
@MaciejPiechotka Well true. But some people call it a trivial field when $0=1$, while others tend to include $0 \neq 1$ in the definition of the field. So it depends... – user17762 Jun 3 '12 at 6:45
Thanks, Marvis. I have written a formal version of your proof. See Lemma 8 beginning at line 312 in – Dan Christensen Jun 4 '12 at 19:38
@DanChristensen Interesting website. I assume it would have involved a lot of work! Good luck! – user17762 Jun 4 '12 at 20:07

$1$ cannot be negative because its sign is also that of $1\cdot 1$, and negative times negative must make positive. Since also $1\ne 0$, it must be positive.

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