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Let $M$ be the set of all rational functions on $\mathbb{R}$. Let product and sum of $f,g\in M$ the continuous continuation of $fg,f+g$ everywhere where these exist.

Show that $$P:=\{\frac{\sum_{j=0}^{n}a_jx^j}{\sum_{j=0}^{m}b_jx^j} \text{ with } a_n,b_m\not= 0|\frac{a_n}{b_m}>0$$ induces an ordering on $M$.

The excercise above is not what I want to solve (yet). I sadly can't find something of "inducing an order", so maybe someone can link to a good explanation, or explain it himself?

Thanks

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Think about the real numbers. $a>b$ if and only if $a-b$ is positive. So, from knowing what the positive reals are, and how subtraction works, we can figure out the ordering of the reals.

That's exactly what's going on here. $P$ induces an ordering given by $$f

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    ok, and why was the ordering not there before P? – 2017-01-21
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    @Socrates Because the ordering is being *defined from* $P$. We declare what our positive elements are going to be, and that then tells us what the ordering must be. We could have gone the other way, of course: declared what our ordering is, and then from that deduced what the positive elements are. But the point is that it's enough to know which elements are positive; in that sense, the set of positive elements *induces* the whole ordering. (More generally, two orderings on a group which respect the group operation are equal if and only if they have the same positive elements.) – 2017-01-21
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    So "inducing an ordering" can be translated to "satisfies the ordering axioms"? – 2017-01-21
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    @Socrates No, that's not right. Rather, it means "provides enough information to determine a unique ordering." The set $P$ itself is not an ordering; an ordering is a set of *ordered pairs* with some properties, and $P$ is just a set of elements. But $P$ tells us how to *build* an ordering: to tell if $a$b-a\in P$. – 2017-01-21
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    Ok, I think I understood now. Thank you for your time. – 2017-01-21