Rational function must have polynomials?

Question

In Oscar Fernandez' Everyday Calculus, he refers to this equation as a rational function: $$r(V) = k{\sqrt{P_0l} \over V}.$$ Don't both the numerator and denominator have to be polynomials to be called a rational function? Is he, maybe, calling it this simply because it contains a ratio of $ \sqrt{P_0l} $ over $ V $?

Answer 1

Fernandez is calling this a rational function, because the numerator and denominator are both polynomials in $V$.

Remember that a polynomail of degree $n$ in $V$ is just an expression of the form:$$\alpha_0+\alpha_1V+\alpha_2V^2+\alpha_3V^3+\ldots+\alpha_nV^n.$$

So if we choose $\alpha_0=k\sqrt{P_0 l}$ and $\alpha_i=0$, for $i=1,2,\ldots,n$, then we see that $k\sqrt{P_0 l}$ is a polynomial in $V$ of degree $0$.

Likwise we can choose $\alpha_0=0$, $\alpha_1=1$ and $\alpha_i=0$, for $i=2,3,\ldots,n$, and we see that $V$ is a polynomial in $V$ of degree $1$.

So the expresion $k\frac{\sqrt{P_0l}}{V}$ is a rational function indeed.