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Let $P:\mathbb{Q}\,{\rightarrow}\,\mathbb{Q}$ and $Q:\mathbb{Q}\,{\rightarrow}\,\mathbb{Q}$ be polynomial functions. Let $S=\mathbb{Q}{\setminus}{\{q\,{\in}\,\mathbb{Q}:Q(q)=0}\}$. Then a function $R:S\,{\rightarrow}\,\mathbb{Q}$, defined as $R(x)=P(x)*[Q(x)]^{-1}$ is called a rational function.

Can $S={\emptyset}$?

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    Only if $Q$ is the zero polynomial. A nonzero polynomial of degree $n$ has at most $n$ roots. – 2017-01-14
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    If $R$ is an infinite integral domain and if $P \in R[X]$ then $P(a)=0 \;\forall a \in R \implies P = 0 \in R[X]$. Apply this to $R=\Bbb Q$. This is false for $R=\Bbb F_2$ (which is finite) or $R=\Bbb F_2^{\Bbb N}$ (which is not a domain), and $P(X)=X^2+X$. – 2017-01-14

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