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

Dan Christensen
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2 Answers2

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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. – Maja Piechotka Jun 03 '12 at 05:19
  • @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... –  Jun 03 '12 at 06:45
  • Thanks, Marvis. I have written a formal version of your proof. See Lemma 8 beginning at line 312 in http://www.dcproof.com/FieldLemmas.htm – Dan Christensen Jun 04 '12 at 19:38
  • @DanChristensen Interesting website. I assume it would have involved a lot of work! Good luck! –  Jun 04 '12 at 20:07
  • Actually, your definition of a total order is incorrect. A total order is merely a partial order that is strongly connected, by definition. That the ring operator $+$ is bi-monotonic is not implied by the definition of a total order, but rather, it is a consequence of how an additive order is defined on a ring. In fact, in an arbitrary ordered field, it is impossible to distinguish a priori a total order from its converse order via the axioms alone. The only way to uniquely specify the order is to specify that $0\lt1$ or $0\gt1,$ one way or the other. – Angel Nov 30 '21 at 15:03
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$1$ 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.

hmakholm left over Monica
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