Natural domain of rational functions

Question

I had an argument about the natural domain of rational functions.

Consider $f(x) = \frac{P(x)}{Q(x)},$ where $P$, and $Q$ are (real) polynomials. Is the natural domain of $f$ the set $\{ x \in \mathbb{R}: Q(x) \neq 0\}$? Or should we first simplify the fraction? For instance, let us consider $P = Q = x$, so that $$f(x) = \frac{x}{x}. $$ I sustain that the domain should exclude $x = 0$, but my friend argues that the functions $f \equiv 1$ on $\mathbb{R}$ thus $0$ belongs to the natural domain of the function. Who is right?

Answer 1

It depends on how you define the function obtained from a polynomial fraction, as well as on the definition of "polynomial fraction". This is a very subtle question, and I'll do my best to provide some ideas, if not give a definitive answer.

First, it is important to distinguish between a rational function, which is a function, and what I'll call a polynomial fraction, which, strictly speaking, isn't. To simplify, let's look at the analogous case of polynomials and polynomial functions. A polynomial is a formal expression like $x^2+2x-3$. It's just a string of variable symbols, numbers, and arithmetic symbols; it isn't (strictly speaking) a function. However, each polynomial $p$ gives rise to a polynomial function $f_p : \Bbb{R} \rightarrow \Bbb{R}$, in this example given by $f_p(x) = x^2+2x-3$. This distinction isn't important in basic algebra and calculus, but it does exist, and it becomes important in more advanced contexts. For instance, in modular arithmetic, you can have different polynomials which give rise to the same polynomial function. An example is that the polynomials $x^3 - 1$ and $x+2$ give rise to the same polynomial function modulo 3 (that is, in $\Bbb{Z}/3\Bbb{Z}$). Different polynomials, same corresponding polynomial function.

Next, let's go back to (what I'm calling) polynomial fractions. One way to define a polynomial fraction is to say it is a formal expression of the form $p(x)/q(x)$, where $p$ and $q$ are polynomials. With this approach, if we want to assign a function to each polynomial fraction, then we are in a sense forced to exclude all zeroes of $q$ from the domain. The corresponding rational function would have domain $\Bbb{R}\backslash\{\text{zeroes of }q\}$, codomain $\Bbb{R}$, and formula $f(x) = p(x)/q(x)$ for all $x$ in the domain.

However, things get tricky when we start asking about reducing fractions. For instance, is the polynomial fraction $x/x$ equal to the polynomial fraction $1$? Or are they merely equivalent? We consider the regular fractions $2/2$ and $1$ to be equal, so should we do the same with polynomial fractions? This is something that is done in more advanced math, giving us the field of fractions $\Bbb{R}(x)$ of the ring of polynomials $\Bbb{R}[x]$, where we really do consider $x/x$ and $1$ to be "equal". However, this means that our naive way of assigning functions to polynomial fractions is no longer valid, because two fractions $x/x$ and $1$ which are equal would give functions that are not equal (because the domains are different). So, this is no longer a valid mapping of polynomial fractions to functions. To fix this, we could say that the function associated to $p(x)/q(x)$ is the one with the largest possible domain.

However, the notion of "largest possible domain" may not be valid. In the case of single-variable real polynomials, we can find a work-around: simply reduce the fraction as much as possible, and only exclude the zeroes of the denominator in the reduced form. But once we move into the realm of multivariable polynomials over a different ring (i.e. a different number system), this reduction algorithm may not exist or may give more than one possible result. So, this operation may no longer be well-defined, and it can be easier to just go back to considering polynomial fractions without this extra equality.

It may also be worth pointing out that we should not conflate "extending by simplifying the fraction" with "extending by continuity". On one hand, we think of $x/x$ as having a removable discontinuity, and so we can extend its domain to include $0$ by continuity. On the other hand, we can think that $x/x = 1$ as polynomial fractions, and the right hand side's domain already includes $0$. We might be tempted to think that there are always these two ways of looking at it, but this is not the case with multivariate polynomials. For example, $f(x,y) = (x^2+y^2)/y$ has a removable discontinuity at $(x,y)=(0,0)$, even though we cannot remove this discontinuity by simplifying the fraction.

Answer 2

Typically, when one says $$f(x)=R(x)/Q(X)$$ is a rational function, they mean that the fraction is in reduced form, i.e. $R(x)$ and $Q(x)$ have no common factor. In such a case, the largest domain that makes sense (what you would call a "natural domain") is $\{x \in \mathbb{R} : Q(x) \neq 0\}$.

Thus, it makes sense to reduce the fraction before thinking about the domain.

Answer 3

There is no ambiguity. If $Q(x) = 0$, then your function is simply undefined, since it is defined by $f(x) = \frac{P(x)}{Q(x)}$. Now if your fraction can be simplified, for instance if $P(x) = x(x-1)$ and $Q(x)= x(x-2)$, then you can extend $f$ by continuity. In my example, the domain of $f$ would be $\Bbb R \setminus \{0, 2\}$ and $f$ could be extended by continuity at $0$, but not at $2$.

Answer 4

To reduce a fraction, we have to divide numerator and denominator by a number not $0$. So $\frac{x}{x}=1$ for $x\neq 0$.