I have the following definition for rational functions:
"Let $p(x), q(x)$ be polynomials and define $D=\{x\in \mathbb{R} : q(x) \neq 0 \}$
Then, the function
$$f:D \rightarrow \mathbb{R}$$
$$f(x) = \frac{p(x)}{q(x)}$$
is said to be rational"
My question is, if you had a proper subset A $\subset$ D and defined function:
$$g:A \rightarrow \mathbb{R}$$
$$g(x) = \frac{p(x)}{q(x)}$$
Would $g$ be considered a rational function, or does the domain not being $D=\{x\in \mathbb{R} : q(x) \neq 0 \}$ mean that the function is not considered rational?
If in the above case $g$ is not considered a rational function, then if we have
$$f_1 : D_1 \rightarrow \mathbb{R}$$ $$f_1(x) = \frac{p_1(x)}{q_1(x)}$$ $$D_1 = \{x\in \mathbb{R} : q_1(x) \neq 0 \}$$ and $$f_2 : D_2 \rightarrow \mathbb{R}$$ $$f_2(x) = \frac{p_2(x)}{q_2(x)}$$ $$D_2 = \{x\in \mathbb{R} : q_2(x) \neq 0 \}$$ Is the function $\frac{f_1}{f_2}$ a rational function? What would its domain be?
If $A\subset\Bbb R$ and $p(x),q(x)\in\Bbb R[x]$ are such that $(\forall y\in A):q(y)\ne0$, then $\frac{p(x)}{q(x)}$ is a rational function. And a rational function (whose domain is a subset of $\Bbb R$) is any function which can be defined this way.