Timeline for Is simplifying a rational function considered as a continuous extension?
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| Dec 30, 2015 at 20:51 | comment | added | quid | Tangentially as the reals are infinite it'd be isomorphic in a natural way to a set of functions/maps, namely the rational ones without removable singularity. And in some sense this is the main point of my answer: it is a convention to interpret the expression by first considering the two polynomials as definining maps and taking quotients then (where possible), yet one could also take the quotient in a formal way first and assign a function/map to the formal quotient. | |
| Dec 30, 2015 at 20:48 | comment | added | quid | I did not say and did not intend to imply it is a set of functions, I merely used "field of rational functions" as name for the algebraic structure (also note that I purposefully used a captial $X$ and at the start of the post said mapping not function), which as you said is the quotient field of the ring of polynomials (in one variable) or equivalently the purely transcendental field extension of transcedence degree $1$. I believe to call it field of rational functions or rational function field (in one variable) is not uncommon. | |
| Dec 30, 2015 at 16:10 | comment | added | vonbrand | The rationals in $\mathbb{R}(X)$ are most definitely not a "set of functions", they are (or can be represented as) a set of quotients of formal polynomials. The equality between such quotients is defined by formal manipulations, not by looking what happens when the indeterminate $X$ takes a value. | |
| Dec 29, 2015 at 16:55 | history | answered | quid | CC BY-SA 3.0 |