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

2 Answers 2

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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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    @DanChristensen Interesting website. I assume it would have involved a lot of work! Good luck!2012-06-04
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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.