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Suppose we have a continuous real valued function $f(x)$ which takes the form of a polynomial for all rationals, i.e. $\exists$ $a_0, a_1, ..., a_n \in \mathbb R$ such that $$f(x)=a_0x^n+a_1x^{n-1}+ \cdots +a_{n-1}x+a_n \space \space \space \space \space \forall x \in \mathbb Q.$$ Now, with the help of continuity, can I conclude that $$f(x)=a_0x^n+a_1x^{n-1}+ \cdots +a_{n-1}x+a_n \space \space \space \space \space \forall x \in \mathbb R.$$ If yes, then how? I came to such a question while solving functional equations where it is possible to find the function for rationals from the given equation easily, whereas for irrationals it is quite difficult.

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Take any real number $x$. Now, take a sequence $q_n$ that converges towards $x$ and look at the limit $$\lim_{k\rightarrow \infty} f(q_k)=\lim_{k\rightarrow\infty} a_0q_k^n + \dots +a_n = a_0x^n + \dots + a_n.$$ By continuity, this limit is equal to $f(x)$, proving what you need.

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