In an ordered field, must the multiplicative identity be positive? Or must it be defined as such?
In an ordered field, must 1 be positive?
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field-theory
2 Answers
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Remember that for a total order,
- If $a \leq b$, then $a+c \leq b+c$.
- 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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1Minor point - $0 = 1$ is excluded by definition of field. – 2012-06-03
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0@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... – 2012-06-03
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0Thanks, Marvis. I have written a formal version of your proof. See Lemma 8 beginning at line 312 in http://www.dcproof.com/FieldLemmas.htm – 2012-06-04
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0@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.