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I am working through a Calculus book and I found an exercise, which I am not able to solve:

Let $K$ be a field of rational functions over $\mathbb{R}$ and let $a,b$ be arbitrary in $\mathbb{Q}$ but not all $b$ are $0$. And let $m,n$ be arbitrary in $\mathbb{N}$. We define $P:=\left.\left\{\frac{\sum_{i=0}^n a_ix^i}{\sum_{i=0}^m b_ix^i}\;\right|\;a_nb_m\lt 0\right\}.$

Is $P$ a prepositive cone?

Does anybody have a hint. My idea was to express $-1$ as a sum of squares but that didn't really work for me.

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    It's a german calculus book "Analysis Band 1, Erhardt Behrends", first chapter :S2011-05-10

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The inequality $a_nb_m<0$ goes the wrong way. Every square in the field should be in the cone, since we are trying to produce an order on the field compatible with the field structure. But it is obvious that for example $1$, also $x^2$, are not expressible as a quotient of the type you mention.

Are you sure that the inequality is not $a_nb_m>0$? That will in fact produce a field ordering.

What's Hilbert's 17th problem doing in a calculus book?

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    @monoid: Yes, $-1$ is fatal. Not getting the squares violates the definition of prepositive cone, but could be an oversight fixable by throwing more stuff in to really make it$a$prepositive cone.2011-05-11